Using the Fund. Theorem of Calc find $h(x)=int_{2}^frac{1}{x}sin^4tdt$'s derivative.












1












$begingroup$


Consider the integral $$h(x)=int_{2}^frac{1}{x}sin^4(t)dt$$



In my notes I have the integral equal to
$$sin^4left(-frac{1}{x^2}right)cdotfrac{1}{x}$$



and the following answer as
$$h'(x)=-sin^4frac{1}{x}/{x^2}$$



Obviously I skipped steps and looked up the answer in the back of the book, would really appreciate it if someone could fill me in.






calculus







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  • $begingroup$
    You should explain also what you have tried so far.
    $endgroup$
    – gimusi
    Dec 10 '18 at 0:38
















1












$begingroup$


Consider the integral $$h(x)=int_{2}^frac{1}{x}sin^4(t)dt$$



In my notes I have the integral equal to
$$sin^4left(-frac{1}{x^2}right)cdotfrac{1}{x}$$



and the following answer as
$$h'(x)=-sin^4frac{1}{x}/{x^2}$$



Obviously I skipped steps and looked up the answer in the back of the book, would really appreciate it if someone could fill me in.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should explain also what you have tried so far.
    $endgroup$
    – gimusi
    Dec 10 '18 at 0:38














1












1








1


2



$begingroup$


Consider the integral $$h(x)=int_{2}^frac{1}{x}sin^4(t)dt$$



In my notes I have the integral equal to
$$sin^4left(-frac{1}{x^2}right)cdotfrac{1}{x}$$



and the following answer as
$$h'(x)=-sin^4frac{1}{x}/{x^2}$$



Obviously I skipped steps and looked up the answer in the back of the book, would really appreciate it if someone could fill me in.






calculus







share|cite|improve this question











$endgroup$




Consider the integral $$h(x)=int_{2}^frac{1}{x}sin^4(t)dt$$



In my notes I have the integral equal to
$$sin^4left(-frac{1}{x^2}right)cdotfrac{1}{x}$$



and the following answer as
$$h'(x)=-sin^4frac{1}{x}/{x^2}$$



Obviously I skipped steps and looked up the answer in the back of the book, would really appreciate it if someone could fill me in.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 0:37









gimusi

92.8k84494




92.8k84494










asked Dec 10 '18 at 0:36









Eric BrownEric Brown

737




737












  • $begingroup$
    You should explain also what you have tried so far.
    $endgroup$
    – gimusi
    Dec 10 '18 at 0:38


















  • $begingroup$
    You should explain also what you have tried so far.
    $endgroup$
    – gimusi
    Dec 10 '18 at 0:38
















$begingroup$
You should explain also what you have tried so far.
$endgroup$
– gimusi
Dec 10 '18 at 0:38




$begingroup$
You should explain also what you have tried so far.
$endgroup$
– gimusi
Dec 10 '18 at 0:38










2 Answers
2






active

oldest

votes


















0












$begingroup$

Set $mathscr{F}'(x) : = sin^4(x)$, i.e., $mathscr{F}$ is the antiderivative of $sin^4(x)$. Then by the fundamental theorem of calculus $$h(x) = mathscr{F} left( frac{1}{x} right) - mathscr{F}left( 2 right).$$ Differentiating this expression, we see that $$h'(x) = -frac{1}{x^2} mathscr{F}' left( frac{1}{x} right).$$ Since $mathscr{F}'(x) = sin^4(x)$, it follows that $$h'(x) = - frac{1}{x^2} sin^4 left( frac{1}{x} right).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful.
    $endgroup$
    – gimusi
    Dec 10 '18 at 0:45












  • $begingroup$
    I agree, but having not studied this topic in a while I had no idea on how to advance.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:48










  • $begingroup$
    @AmorFati Also how does $F(2)$ become $frac{1}{x}$?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:51












  • $begingroup$
    @gimusi I feel this is sufficiently elementary to offer a complete solution.
    $endgroup$
    – AmorFati
    Dec 10 '18 at 0:51






  • 1




    $begingroup$
    @AmorFati Ahhh I see now, thanks for your help.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:52



















0












$begingroup$

Say we had a function $F$ so that $F'(t)=sin^4t$. Then $h(x)=F(1/x)-F(2)$, so
$$h'(x) =^* F'left(frac1xright)left(-frac1{x^2}right)=-frac1{x^2}left(sin^4left(frac1xright)right).$$
Note that the expression after $=^*$ is derived by applying both the chain rule and the power rule. The chain rule comes in as $frac d{dx}u(v(x))=u'(v(x))v'(x)$ where $u(v)=F(v)$, and $v(x)=1/x$. The power rule comes in when calculating $v'(x)=(x^{-1})'=-1x^{-1-1}=-1/x^2$.






share|cite|improve this answer









$endgroup$













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    2 Answers
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    2 Answers
    2






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    0












    $begingroup$

    Set $mathscr{F}'(x) : = sin^4(x)$, i.e., $mathscr{F}$ is the antiderivative of $sin^4(x)$. Then by the fundamental theorem of calculus $$h(x) = mathscr{F} left( frac{1}{x} right) - mathscr{F}left( 2 right).$$ Differentiating this expression, we see that $$h'(x) = -frac{1}{x^2} mathscr{F}' left( frac{1}{x} right).$$ Since $mathscr{F}'(x) = sin^4(x)$, it follows that $$h'(x) = - frac{1}{x^2} sin^4 left( frac{1}{x} right).$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful.
      $endgroup$
      – gimusi
      Dec 10 '18 at 0:45












    • $begingroup$
      I agree, but having not studied this topic in a while I had no idea on how to advance.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:48










    • $begingroup$
      @AmorFati Also how does $F(2)$ become $frac{1}{x}$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:51












    • $begingroup$
      @gimusi I feel this is sufficiently elementary to offer a complete solution.
      $endgroup$
      – AmorFati
      Dec 10 '18 at 0:51






    • 1




      $begingroup$
      @AmorFati Ahhh I see now, thanks for your help.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:52
















    0












    $begingroup$

    Set $mathscr{F}'(x) : = sin^4(x)$, i.e., $mathscr{F}$ is the antiderivative of $sin^4(x)$. Then by the fundamental theorem of calculus $$h(x) = mathscr{F} left( frac{1}{x} right) - mathscr{F}left( 2 right).$$ Differentiating this expression, we see that $$h'(x) = -frac{1}{x^2} mathscr{F}' left( frac{1}{x} right).$$ Since $mathscr{F}'(x) = sin^4(x)$, it follows that $$h'(x) = - frac{1}{x^2} sin^4 left( frac{1}{x} right).$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful.
      $endgroup$
      – gimusi
      Dec 10 '18 at 0:45












    • $begingroup$
      I agree, but having not studied this topic in a while I had no idea on how to advance.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:48










    • $begingroup$
      @AmorFati Also how does $F(2)$ become $frac{1}{x}$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:51












    • $begingroup$
      @gimusi I feel this is sufficiently elementary to offer a complete solution.
      $endgroup$
      – AmorFati
      Dec 10 '18 at 0:51






    • 1




      $begingroup$
      @AmorFati Ahhh I see now, thanks for your help.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:52














    0












    0








    0





    $begingroup$

    Set $mathscr{F}'(x) : = sin^4(x)$, i.e., $mathscr{F}$ is the antiderivative of $sin^4(x)$. Then by the fundamental theorem of calculus $$h(x) = mathscr{F} left( frac{1}{x} right) - mathscr{F}left( 2 right).$$ Differentiating this expression, we see that $$h'(x) = -frac{1}{x^2} mathscr{F}' left( frac{1}{x} right).$$ Since $mathscr{F}'(x) = sin^4(x)$, it follows that $$h'(x) = - frac{1}{x^2} sin^4 left( frac{1}{x} right).$$






    share|cite|improve this answer









    $endgroup$



    Set $mathscr{F}'(x) : = sin^4(x)$, i.e., $mathscr{F}$ is the antiderivative of $sin^4(x)$. Then by the fundamental theorem of calculus $$h(x) = mathscr{F} left( frac{1}{x} right) - mathscr{F}left( 2 right).$$ Differentiating this expression, we see that $$h'(x) = -frac{1}{x^2} mathscr{F}' left( frac{1}{x} right).$$ Since $mathscr{F}'(x) = sin^4(x)$, it follows that $$h'(x) = - frac{1}{x^2} sin^4 left( frac{1}{x} right).$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 0:41









    AmorFatiAmorFati

    406529




    406529












    • $begingroup$
      As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful.
      $endgroup$
      – gimusi
      Dec 10 '18 at 0:45












    • $begingroup$
      I agree, but having not studied this topic in a while I had no idea on how to advance.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:48










    • $begingroup$
      @AmorFati Also how does $F(2)$ become $frac{1}{x}$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:51












    • $begingroup$
      @gimusi I feel this is sufficiently elementary to offer a complete solution.
      $endgroup$
      – AmorFati
      Dec 10 '18 at 0:51






    • 1




      $begingroup$
      @AmorFati Ahhh I see now, thanks for your help.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:52


















    • $begingroup$
      As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful.
      $endgroup$
      – gimusi
      Dec 10 '18 at 0:45












    • $begingroup$
      I agree, but having not studied this topic in a while I had no idea on how to advance.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:48










    • $begingroup$
      @AmorFati Also how does $F(2)$ become $frac{1}{x}$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:51












    • $begingroup$
      @gimusi I feel this is sufficiently elementary to offer a complete solution.
      $endgroup$
      – AmorFati
      Dec 10 '18 at 0:51






    • 1




      $begingroup$
      @AmorFati Ahhh I see now, thanks for your help.
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 0:52
















    $begingroup$
    As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful.
    $endgroup$
    – gimusi
    Dec 10 '18 at 0:45






    $begingroup$
    As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful.
    $endgroup$
    – gimusi
    Dec 10 '18 at 0:45














    $begingroup$
    I agree, but having not studied this topic in a while I had no idea on how to advance.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:48




    $begingroup$
    I agree, but having not studied this topic in a while I had no idea on how to advance.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:48












    $begingroup$
    @AmorFati Also how does $F(2)$ become $frac{1}{x}$?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:51






    $begingroup$
    @AmorFati Also how does $F(2)$ become $frac{1}{x}$?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:51














    $begingroup$
    @gimusi I feel this is sufficiently elementary to offer a complete solution.
    $endgroup$
    – AmorFati
    Dec 10 '18 at 0:51




    $begingroup$
    @gimusi I feel this is sufficiently elementary to offer a complete solution.
    $endgroup$
    – AmorFati
    Dec 10 '18 at 0:51




    1




    1




    $begingroup$
    @AmorFati Ahhh I see now, thanks for your help.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:52




    $begingroup$
    @AmorFati Ahhh I see now, thanks for your help.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 0:52











    0












    $begingroup$

    Say we had a function $F$ so that $F'(t)=sin^4t$. Then $h(x)=F(1/x)-F(2)$, so
    $$h'(x) =^* F'left(frac1xright)left(-frac1{x^2}right)=-frac1{x^2}left(sin^4left(frac1xright)right).$$
    Note that the expression after $=^*$ is derived by applying both the chain rule and the power rule. The chain rule comes in as $frac d{dx}u(v(x))=u'(v(x))v'(x)$ where $u(v)=F(v)$, and $v(x)=1/x$. The power rule comes in when calculating $v'(x)=(x^{-1})'=-1x^{-1-1}=-1/x^2$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Say we had a function $F$ so that $F'(t)=sin^4t$. Then $h(x)=F(1/x)-F(2)$, so
      $$h'(x) =^* F'left(frac1xright)left(-frac1{x^2}right)=-frac1{x^2}left(sin^4left(frac1xright)right).$$
      Note that the expression after $=^*$ is derived by applying both the chain rule and the power rule. The chain rule comes in as $frac d{dx}u(v(x))=u'(v(x))v'(x)$ where $u(v)=F(v)$, and $v(x)=1/x$. The power rule comes in when calculating $v'(x)=(x^{-1})'=-1x^{-1-1}=-1/x^2$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Say we had a function $F$ so that $F'(t)=sin^4t$. Then $h(x)=F(1/x)-F(2)$, so
        $$h'(x) =^* F'left(frac1xright)left(-frac1{x^2}right)=-frac1{x^2}left(sin^4left(frac1xright)right).$$
        Note that the expression after $=^*$ is derived by applying both the chain rule and the power rule. The chain rule comes in as $frac d{dx}u(v(x))=u'(v(x))v'(x)$ where $u(v)=F(v)$, and $v(x)=1/x$. The power rule comes in when calculating $v'(x)=(x^{-1})'=-1x^{-1-1}=-1/x^2$.






        share|cite|improve this answer









        $endgroup$



        Say we had a function $F$ so that $F'(t)=sin^4t$. Then $h(x)=F(1/x)-F(2)$, so
        $$h'(x) =^* F'left(frac1xright)left(-frac1{x^2}right)=-frac1{x^2}left(sin^4left(frac1xright)right).$$
        Note that the expression after $=^*$ is derived by applying both the chain rule and the power rule. The chain rule comes in as $frac d{dx}u(v(x))=u'(v(x))v'(x)$ where $u(v)=F(v)$, and $v(x)=1/x$. The power rule comes in when calculating $v'(x)=(x^{-1})'=-1x^{-1-1}=-1/x^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 0:48









        YiFanYiFan

        3,7841627




        3,7841627






























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