Differentiation for a function in the integral form.












2












$begingroup$


I want to know generally how we differentiate a function $F(x)$ in the following form,
$$F(x)=int_a^x f(x,t)dt$$
For example, if we can work out the explicit form of $F(x)$ as the example below $$F(x)=int_0^x e^{xt}dt=frac{e^{x^2}-1}{x}$$
if we differentiate $F(x)$, we get
$$frac {dF(x)}{dx}=frac{2x^2e^{x^2}-e^{x^2}+1}{x^2}$$
This simple example shows a very different approach from the case when we different $F(x)=int_a^xf(t)dt$, which simply gives us $f(x)$.



Now suppose I have the formula for calculating the money I get after $T$ years, $P(T)$ with continuous depositing, of which the rate is given by $S(t)$, and continuous compounding, of which the annual rate is given by $r$. The formula is $$P(T)=int_0^TS(t)e^{r(T-t)}dt$$
Since the $S(t)$ is not given explicitly, then how do I get an expression for $frac{dP(T)}{dT}$?



I have tried using the first principle by
$$frac{dP(T)}{dT}=lim_{Delta Tto0} frac{P(T+Delta T)-P(T)}{Delta T}$$ and I could not figure it out. Is there a way to find an expression for the derivative $frac{dP(T)}{dT}$?






calculus derivatives implicit-differentiation







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I want to know generally how we differentiate a function $F(x)$ in the following form,
    $$F(x)=int_a^x f(x,t)dt$$
    For example, if we can work out the explicit form of $F(x)$ as the example below $$F(x)=int_0^x e^{xt}dt=frac{e^{x^2}-1}{x}$$
    if we differentiate $F(x)$, we get
    $$frac {dF(x)}{dx}=frac{2x^2e^{x^2}-e^{x^2}+1}{x^2}$$
    This simple example shows a very different approach from the case when we different $F(x)=int_a^xf(t)dt$, which simply gives us $f(x)$.



    Now suppose I have the formula for calculating the money I get after $T$ years, $P(T)$ with continuous depositing, of which the rate is given by $S(t)$, and continuous compounding, of which the annual rate is given by $r$. The formula is $$P(T)=int_0^TS(t)e^{r(T-t)}dt$$
    Since the $S(t)$ is not given explicitly, then how do I get an expression for $frac{dP(T)}{dT}$?



    I have tried using the first principle by
    $$frac{dP(T)}{dT}=lim_{Delta Tto0} frac{P(T+Delta T)-P(T)}{Delta T}$$ and I could not figure it out. Is there a way to find an expression for the derivative $frac{dP(T)}{dT}$?






    calculus derivatives implicit-differentiation







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I want to know generally how we differentiate a function $F(x)$ in the following form,
      $$F(x)=int_a^x f(x,t)dt$$
      For example, if we can work out the explicit form of $F(x)$ as the example below $$F(x)=int_0^x e^{xt}dt=frac{e^{x^2}-1}{x}$$
      if we differentiate $F(x)$, we get
      $$frac {dF(x)}{dx}=frac{2x^2e^{x^2}-e^{x^2}+1}{x^2}$$
      This simple example shows a very different approach from the case when we different $F(x)=int_a^xf(t)dt$, which simply gives us $f(x)$.



      Now suppose I have the formula for calculating the money I get after $T$ years, $P(T)$ with continuous depositing, of which the rate is given by $S(t)$, and continuous compounding, of which the annual rate is given by $r$. The formula is $$P(T)=int_0^TS(t)e^{r(T-t)}dt$$
      Since the $S(t)$ is not given explicitly, then how do I get an expression for $frac{dP(T)}{dT}$?



      I have tried using the first principle by
      $$frac{dP(T)}{dT}=lim_{Delta Tto0} frac{P(T+Delta T)-P(T)}{Delta T}$$ and I could not figure it out. Is there a way to find an expression for the derivative $frac{dP(T)}{dT}$?






      calculus derivatives implicit-differentiation







      share|cite|improve this question









      $endgroup$




      I want to know generally how we differentiate a function $F(x)$ in the following form,
      $$F(x)=int_a^x f(x,t)dt$$
      For example, if we can work out the explicit form of $F(x)$ as the example below $$F(x)=int_0^x e^{xt}dt=frac{e^{x^2}-1}{x}$$
      if we differentiate $F(x)$, we get
      $$frac {dF(x)}{dx}=frac{2x^2e^{x^2}-e^{x^2}+1}{x^2}$$
      This simple example shows a very different approach from the case when we different $F(x)=int_a^xf(t)dt$, which simply gives us $f(x)$.



      Now suppose I have the formula for calculating the money I get after $T$ years, $P(T)$ with continuous depositing, of which the rate is given by $S(t)$, and continuous compounding, of which the annual rate is given by $r$. The formula is $$P(T)=int_0^TS(t)e^{r(T-t)}dt$$
      Since the $S(t)$ is not given explicitly, then how do I get an expression for $frac{dP(T)}{dT}$?



      I have tried using the first principle by
      $$frac{dP(T)}{dT}=lim_{Delta Tto0} frac{P(T+Delta T)-P(T)}{Delta T}$$ and I could not figure it out. Is there a way to find an expression for the derivative $frac{dP(T)}{dT}$?






      calculus derivatives implicit-differentiation




      calculus derivatives implicit-differentiation






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      share|cite|improve this question











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      asked Dec 5 '18 at 9:39









      Kenneth NyeKenneth Nye

      19210




      19210






















          3 Answers
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          1












          $begingroup$

          We have that
          $$
          eqalign{
          & F(x + dx) - F(x) = int_{t = a}^{x + dx} {f(x + dx,t)dt} - int_{t = a}^x {f(x,t)dt} = cr
          & = int_{t = a}^x {left( {f(x + dx,t) - f(x,t)} right)dt} + int_{t = x}^{x + dx} {f(x + dx,t)dt} = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x + dx,x)dx = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x,x)dx + f_x (x,x)left( {dx} right)^2 quad Rightarrow cr }$$




          $$ Rightarrow quad {d over {dx}}F(x) = int_{t = a}^x {{partial over {partial x}}f(x,t)dt} + f(x,x)
          $$




          So in your example
          $$
          eqalign{
          & {d over {dx}}F(x) = int_{t = 0}^x {{partial over {partial x}}e^{,x,t} dt} + e^{,x^{,2} } = cr
          & = int_{t = 0}^x {te^{,x,t} dt} + e^{,x^{,2} } = left. {{{e^{,x,t} left( {tx - 1} right)} over {x^2 }},} right|_{t = 0}^x + e^{,x^{,2} } = cr
          & = {{e^{,x^{,2} } left( {,x^{,2} - 1} right)} over {x^2 }} + {1 over {x^2 }} + e^{,x^{,2} }
          = {{,2e^{,x^{,2} } x^{,2} - e^{,x^{,2} } + 1} over {x^2 }} cr}
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for providing the detailed but simply proof for Leibniz integral rule. I will stop the discussion at your answer since my question was totally solved by this rule.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 11:12



















          0












          $begingroup$

          You can impose $a=Tt$ so you have that the integral is



          $P(T)=frac{1}{T}int_0^1 S(frac{a}{T})e^{rfrac{(T^2-a)}{T}}da$



          So



          $frac{d}{dT}P(T)=frac{1}{T}int_0^1[-frac{a}{T^2}S’(frac{a}{T})e^{rfrac{(T^2-a)}{T}}+$



          $+r(1+frac{a}{T^2}) e^{rfrac{(T^2-a)}{T}} S(frac{a}{T})]da$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I get what you mean, basically try to get rid of variables for the limits, but I think your substitution result is not right.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:54










          • $begingroup$
            Now I think it is right
            $endgroup$
            – Federico Fallucca
            Dec 5 '18 at 9:55










          • $begingroup$
            Is it generally true you can simply swap differentiation and integration when the limits are constant? Can you refer me a theorem? Thanks
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:56






          • 1




            $begingroup$
            I realize it is the lebiz integral rule. en.m.wikipedia.org/wiki/Leibniz_integral_rule I need to upgrade my knowledge.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:33



















          0












          $begingroup$

          You want to use Leibniz integral rule. If $f(x,t)$ is regular enough the following formula holds



          $$frac{d}{dx} int_{a(x)}^{b(x)} f(x,t), dt = fbig(x,b(x)big) b'(x) - fbig(x,a(x)big)a'(x) + int_{a(x)}^{b(x)}frac{partial}{partial x}f(x,t), dt,.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Just now i searched and already found this formula on wiki. I need to upgrade my knowledge. Currently I only have math knowledge at uni year 3 level. Hahaha.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:41











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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          We have that
          $$
          eqalign{
          & F(x + dx) - F(x) = int_{t = a}^{x + dx} {f(x + dx,t)dt} - int_{t = a}^x {f(x,t)dt} = cr
          & = int_{t = a}^x {left( {f(x + dx,t) - f(x,t)} right)dt} + int_{t = x}^{x + dx} {f(x + dx,t)dt} = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x + dx,x)dx = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x,x)dx + f_x (x,x)left( {dx} right)^2 quad Rightarrow cr }$$




          $$ Rightarrow quad {d over {dx}}F(x) = int_{t = a}^x {{partial over {partial x}}f(x,t)dt} + f(x,x)
          $$




          So in your example
          $$
          eqalign{
          & {d over {dx}}F(x) = int_{t = 0}^x {{partial over {partial x}}e^{,x,t} dt} + e^{,x^{,2} } = cr
          & = int_{t = 0}^x {te^{,x,t} dt} + e^{,x^{,2} } = left. {{{e^{,x,t} left( {tx - 1} right)} over {x^2 }},} right|_{t = 0}^x + e^{,x^{,2} } = cr
          & = {{e^{,x^{,2} } left( {,x^{,2} - 1} right)} over {x^2 }} + {1 over {x^2 }} + e^{,x^{,2} }
          = {{,2e^{,x^{,2} } x^{,2} - e^{,x^{,2} } + 1} over {x^2 }} cr}
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for providing the detailed but simply proof for Leibniz integral rule. I will stop the discussion at your answer since my question was totally solved by this rule.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 11:12
















          1












          $begingroup$

          We have that
          $$
          eqalign{
          & F(x + dx) - F(x) = int_{t = a}^{x + dx} {f(x + dx,t)dt} - int_{t = a}^x {f(x,t)dt} = cr
          & = int_{t = a}^x {left( {f(x + dx,t) - f(x,t)} right)dt} + int_{t = x}^{x + dx} {f(x + dx,t)dt} = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x + dx,x)dx = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x,x)dx + f_x (x,x)left( {dx} right)^2 quad Rightarrow cr }$$




          $$ Rightarrow quad {d over {dx}}F(x) = int_{t = a}^x {{partial over {partial x}}f(x,t)dt} + f(x,x)
          $$




          So in your example
          $$
          eqalign{
          & {d over {dx}}F(x) = int_{t = 0}^x {{partial over {partial x}}e^{,x,t} dt} + e^{,x^{,2} } = cr
          & = int_{t = 0}^x {te^{,x,t} dt} + e^{,x^{,2} } = left. {{{e^{,x,t} left( {tx - 1} right)} over {x^2 }},} right|_{t = 0}^x + e^{,x^{,2} } = cr
          & = {{e^{,x^{,2} } left( {,x^{,2} - 1} right)} over {x^2 }} + {1 over {x^2 }} + e^{,x^{,2} }
          = {{,2e^{,x^{,2} } x^{,2} - e^{,x^{,2} } + 1} over {x^2 }} cr}
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for providing the detailed but simply proof for Leibniz integral rule. I will stop the discussion at your answer since my question was totally solved by this rule.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 11:12














          1












          1








          1





          $begingroup$

          We have that
          $$
          eqalign{
          & F(x + dx) - F(x) = int_{t = a}^{x + dx} {f(x + dx,t)dt} - int_{t = a}^x {f(x,t)dt} = cr
          & = int_{t = a}^x {left( {f(x + dx,t) - f(x,t)} right)dt} + int_{t = x}^{x + dx} {f(x + dx,t)dt} = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x + dx,x)dx = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x,x)dx + f_x (x,x)left( {dx} right)^2 quad Rightarrow cr }$$




          $$ Rightarrow quad {d over {dx}}F(x) = int_{t = a}^x {{partial over {partial x}}f(x,t)dt} + f(x,x)
          $$




          So in your example
          $$
          eqalign{
          & {d over {dx}}F(x) = int_{t = 0}^x {{partial over {partial x}}e^{,x,t} dt} + e^{,x^{,2} } = cr
          & = int_{t = 0}^x {te^{,x,t} dt} + e^{,x^{,2} } = left. {{{e^{,x,t} left( {tx - 1} right)} over {x^2 }},} right|_{t = 0}^x + e^{,x^{,2} } = cr
          & = {{e^{,x^{,2} } left( {,x^{,2} - 1} right)} over {x^2 }} + {1 over {x^2 }} + e^{,x^{,2} }
          = {{,2e^{,x^{,2} } x^{,2} - e^{,x^{,2} } + 1} over {x^2 }} cr}
          $$






          share|cite|improve this answer











          $endgroup$



          We have that
          $$
          eqalign{
          & F(x + dx) - F(x) = int_{t = a}^{x + dx} {f(x + dx,t)dt} - int_{t = a}^x {f(x,t)dt} = cr
          & = int_{t = a}^x {left( {f(x + dx,t) - f(x,t)} right)dt} + int_{t = x}^{x + dx} {f(x + dx,t)dt} = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x + dx,x)dx = cr
          & = left( {int_{t = a}^x {f_x (x,t)dt} } right)dx + f(x,x)dx + f_x (x,x)left( {dx} right)^2 quad Rightarrow cr }$$




          $$ Rightarrow quad {d over {dx}}F(x) = int_{t = a}^x {{partial over {partial x}}f(x,t)dt} + f(x,x)
          $$




          So in your example
          $$
          eqalign{
          & {d over {dx}}F(x) = int_{t = 0}^x {{partial over {partial x}}e^{,x,t} dt} + e^{,x^{,2} } = cr
          & = int_{t = 0}^x {te^{,x,t} dt} + e^{,x^{,2} } = left. {{{e^{,x,t} left( {tx - 1} right)} over {x^2 }},} right|_{t = 0}^x + e^{,x^{,2} } = cr
          & = {{e^{,x^{,2} } left( {,x^{,2} - 1} right)} over {x^2 }} + {1 over {x^2 }} + e^{,x^{,2} }
          = {{,2e^{,x^{,2} } x^{,2} - e^{,x^{,2} } + 1} over {x^2 }} cr}
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 10:51

























          answered Dec 5 '18 at 10:45









          G CabG Cab

          18.8k31238




          18.8k31238












          • $begingroup$
            Thanks for providing the detailed but simply proof for Leibniz integral rule. I will stop the discussion at your answer since my question was totally solved by this rule.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 11:12


















          • $begingroup$
            Thanks for providing the detailed but simply proof for Leibniz integral rule. I will stop the discussion at your answer since my question was totally solved by this rule.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 11:12
















          $begingroup$
          Thanks for providing the detailed but simply proof for Leibniz integral rule. I will stop the discussion at your answer since my question was totally solved by this rule.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 11:12




          $begingroup$
          Thanks for providing the detailed but simply proof for Leibniz integral rule. I will stop the discussion at your answer since my question was totally solved by this rule.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 11:12











          0












          $begingroup$

          You can impose $a=Tt$ so you have that the integral is



          $P(T)=frac{1}{T}int_0^1 S(frac{a}{T})e^{rfrac{(T^2-a)}{T}}da$



          So



          $frac{d}{dT}P(T)=frac{1}{T}int_0^1[-frac{a}{T^2}S’(frac{a}{T})e^{rfrac{(T^2-a)}{T}}+$



          $+r(1+frac{a}{T^2}) e^{rfrac{(T^2-a)}{T}} S(frac{a}{T})]da$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I get what you mean, basically try to get rid of variables for the limits, but I think your substitution result is not right.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:54










          • $begingroup$
            Now I think it is right
            $endgroup$
            – Federico Fallucca
            Dec 5 '18 at 9:55










          • $begingroup$
            Is it generally true you can simply swap differentiation and integration when the limits are constant? Can you refer me a theorem? Thanks
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:56






          • 1




            $begingroup$
            I realize it is the lebiz integral rule. en.m.wikipedia.org/wiki/Leibniz_integral_rule I need to upgrade my knowledge.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:33
















          0












          $begingroup$

          You can impose $a=Tt$ so you have that the integral is



          $P(T)=frac{1}{T}int_0^1 S(frac{a}{T})e^{rfrac{(T^2-a)}{T}}da$



          So



          $frac{d}{dT}P(T)=frac{1}{T}int_0^1[-frac{a}{T^2}S’(frac{a}{T})e^{rfrac{(T^2-a)}{T}}+$



          $+r(1+frac{a}{T^2}) e^{rfrac{(T^2-a)}{T}} S(frac{a}{T})]da$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I get what you mean, basically try to get rid of variables for the limits, but I think your substitution result is not right.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:54










          • $begingroup$
            Now I think it is right
            $endgroup$
            – Federico Fallucca
            Dec 5 '18 at 9:55










          • $begingroup$
            Is it generally true you can simply swap differentiation and integration when the limits are constant? Can you refer me a theorem? Thanks
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:56






          • 1




            $begingroup$
            I realize it is the lebiz integral rule. en.m.wikipedia.org/wiki/Leibniz_integral_rule I need to upgrade my knowledge.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:33














          0












          0








          0





          $begingroup$

          You can impose $a=Tt$ so you have that the integral is



          $P(T)=frac{1}{T}int_0^1 S(frac{a}{T})e^{rfrac{(T^2-a)}{T}}da$



          So



          $frac{d}{dT}P(T)=frac{1}{T}int_0^1[-frac{a}{T^2}S’(frac{a}{T})e^{rfrac{(T^2-a)}{T}}+$



          $+r(1+frac{a}{T^2}) e^{rfrac{(T^2-a)}{T}} S(frac{a}{T})]da$






          share|cite|improve this answer











          $endgroup$



          You can impose $a=Tt$ so you have that the integral is



          $P(T)=frac{1}{T}int_0^1 S(frac{a}{T})e^{rfrac{(T^2-a)}{T}}da$



          So



          $frac{d}{dT}P(T)=frac{1}{T}int_0^1[-frac{a}{T^2}S’(frac{a}{T})e^{rfrac{(T^2-a)}{T}}+$



          $+r(1+frac{a}{T^2}) e^{rfrac{(T^2-a)}{T}} S(frac{a}{T})]da$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 10:00

























          answered Dec 5 '18 at 9:47









          Federico FalluccaFederico Fallucca

          1,90219




          1,90219












          • $begingroup$
            I get what you mean, basically try to get rid of variables for the limits, but I think your substitution result is not right.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:54










          • $begingroup$
            Now I think it is right
            $endgroup$
            – Federico Fallucca
            Dec 5 '18 at 9:55










          • $begingroup$
            Is it generally true you can simply swap differentiation and integration when the limits are constant? Can you refer me a theorem? Thanks
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:56






          • 1




            $begingroup$
            I realize it is the lebiz integral rule. en.m.wikipedia.org/wiki/Leibniz_integral_rule I need to upgrade my knowledge.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:33


















          • $begingroup$
            I get what you mean, basically try to get rid of variables for the limits, but I think your substitution result is not right.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:54










          • $begingroup$
            Now I think it is right
            $endgroup$
            – Federico Fallucca
            Dec 5 '18 at 9:55










          • $begingroup$
            Is it generally true you can simply swap differentiation and integration when the limits are constant? Can you refer me a theorem? Thanks
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 9:56






          • 1




            $begingroup$
            I realize it is the lebiz integral rule. en.m.wikipedia.org/wiki/Leibniz_integral_rule I need to upgrade my knowledge.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:33
















          $begingroup$
          I get what you mean, basically try to get rid of variables for the limits, but I think your substitution result is not right.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 9:54




          $begingroup$
          I get what you mean, basically try to get rid of variables for the limits, but I think your substitution result is not right.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 9:54












          $begingroup$
          Now I think it is right
          $endgroup$
          – Federico Fallucca
          Dec 5 '18 at 9:55




          $begingroup$
          Now I think it is right
          $endgroup$
          – Federico Fallucca
          Dec 5 '18 at 9:55












          $begingroup$
          Is it generally true you can simply swap differentiation and integration when the limits are constant? Can you refer me a theorem? Thanks
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 9:56




          $begingroup$
          Is it generally true you can simply swap differentiation and integration when the limits are constant? Can you refer me a theorem? Thanks
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 9:56




          1




          1




          $begingroup$
          I realize it is the lebiz integral rule. en.m.wikipedia.org/wiki/Leibniz_integral_rule I need to upgrade my knowledge.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 10:33




          $begingroup$
          I realize it is the lebiz integral rule. en.m.wikipedia.org/wiki/Leibniz_integral_rule I need to upgrade my knowledge.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 10:33











          0












          $begingroup$

          You want to use Leibniz integral rule. If $f(x,t)$ is regular enough the following formula holds



          $$frac{d}{dx} int_{a(x)}^{b(x)} f(x,t), dt = fbig(x,b(x)big) b'(x) - fbig(x,a(x)big)a'(x) + int_{a(x)}^{b(x)}frac{partial}{partial x}f(x,t), dt,.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Just now i searched and already found this formula on wiki. I need to upgrade my knowledge. Currently I only have math knowledge at uni year 3 level. Hahaha.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:41
















          0












          $begingroup$

          You want to use Leibniz integral rule. If $f(x,t)$ is regular enough the following formula holds



          $$frac{d}{dx} int_{a(x)}^{b(x)} f(x,t), dt = fbig(x,b(x)big) b'(x) - fbig(x,a(x)big)a'(x) + int_{a(x)}^{b(x)}frac{partial}{partial x}f(x,t), dt,.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Just now i searched and already found this formula on wiki. I need to upgrade my knowledge. Currently I only have math knowledge at uni year 3 level. Hahaha.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:41














          0












          0








          0





          $begingroup$

          You want to use Leibniz integral rule. If $f(x,t)$ is regular enough the following formula holds



          $$frac{d}{dx} int_{a(x)}^{b(x)} f(x,t), dt = fbig(x,b(x)big) b'(x) - fbig(x,a(x)big)a'(x) + int_{a(x)}^{b(x)}frac{partial}{partial x}f(x,t), dt,.$$






          share|cite|improve this answer









          $endgroup$



          You want to use Leibniz integral rule. If $f(x,t)$ is regular enough the following formula holds



          $$frac{d}{dx} int_{a(x)}^{b(x)} f(x,t), dt = fbig(x,b(x)big) b'(x) - fbig(x,a(x)big)a'(x) + int_{a(x)}^{b(x)}frac{partial}{partial x}f(x,t), dt,.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 10:38









          Paolo IntuitoPaolo Intuito

          953318




          953318












          • $begingroup$
            Thanks. Just now i searched and already found this formula on wiki. I need to upgrade my knowledge. Currently I only have math knowledge at uni year 3 level. Hahaha.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:41


















          • $begingroup$
            Thanks. Just now i searched and already found this formula on wiki. I need to upgrade my knowledge. Currently I only have math knowledge at uni year 3 level. Hahaha.
            $endgroup$
            – Kenneth Nye
            Dec 5 '18 at 10:41
















          $begingroup$
          Thanks. Just now i searched and already found this formula on wiki. I need to upgrade my knowledge. Currently I only have math knowledge at uni year 3 level. Hahaha.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 10:41




          $begingroup$
          Thanks. Just now i searched and already found this formula on wiki. I need to upgrade my knowledge. Currently I only have math knowledge at uni year 3 level. Hahaha.
          $endgroup$
          – Kenneth Nye
          Dec 5 '18 at 10:41


















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