Find $limlimits_{bto a}frac {1}{b-a}lnleft[frac{a(b-x)}{b(a-x)}right]$ if $x$ is constant using l'Hospital's...
$begingroup$
If $x$ is a constant what do I differentiate with respect to?
My best guess would be $b$. However, is this correct?
Also how do you differentiate that function with respect to $b?$
Do you have to use the product rule, chain rule and quotient rule?
calculus limits
edited Dec 21 '18 at 11:38
user376343
3,9584829
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
$endgroup$
add a comment |
$begingroup$
If $x$ is a constant what do I differentiate with respect to?
My best guess would be $b$. However, is this correct?
Also how do you differentiate that function with respect to $b?$
Do you have to use the product rule, chain rule and quotient rule?
calculus limits
edited Dec 21 '18 at 11:38
user376343
3,9584829
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
$endgroup$
1
$begingroup$
It is correct. Good luck.
$endgroup$
– Galc127
Jan 19 '15 at 14:43
add a comment |
$begingroup$
If $x$ is a constant what do I differentiate with respect to?
My best guess would be $b$. However, is this correct?
Also how do you differentiate that function with respect to $b?$
Do you have to use the product rule, chain rule and quotient rule?
calculus limits
edited Dec 21 '18 at 11:38
user376343
3,9584829
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
$endgroup$
If $x$ is a constant what do I differentiate with respect to?
My best guess would be $b$. However, is this correct?
Also how do you differentiate that function with respect to $b?$
Do you have to use the product rule, chain rule and quotient rule?
calculus limits
calculus limits
edited Dec 21 '18 at 11:38
user376343
3,9584829
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
edited Dec 21 '18 at 11:38
user376343
3,9584829
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
edited Dec 21 '18 at 11:38
user376343
3,9584829
edited Dec 21 '18 at 11:38
user376343
3,9584829
edited Dec 21 '18 at 11:38
user376343
3,9584829
3,9584829
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
asked Jan 19 '15 at 14:39
RobChemRobChem
514412
514412
1
$begingroup$
It is correct. Good luck.
$endgroup$
– Galc127
Jan 19 '15 at 14:43
add a comment |
1
$begingroup$
It is correct. Good luck.
$endgroup$
– Galc127
Jan 19 '15 at 14:43
1
1
$begingroup$
It is correct. Good luck.
$endgroup$
– Galc127
Jan 19 '15 at 14:43
$begingroup$
It is correct. Good luck.
$endgroup$
– Galc127
Jan 19 '15 at 14:43
add a comment |
2 Answers
2
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$begingroup$
Sometimes changing the notation makes it easier. Another way of verifying what you said is replace $b$ by $x$, leave $a$ alone and replace $x$ by $c$. Then differentiate wrt $x$ etc.
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
$endgroup$
add a comment |
$begingroup$
hint: differentiating $lnleft(frac{a(b-x)}{b(a-x)}right)$ with respect to $b$ we get
$$frac{x}{b(b-x)}$$
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Sometimes changing the notation makes it easier. Another way of verifying what you said is replace $b$ by $x$, leave $a$ alone and replace $x$ by $c$. Then differentiate wrt $x$ etc.
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
$endgroup$
add a comment |
$begingroup$
Sometimes changing the notation makes it easier. Another way of verifying what you said is replace $b$ by $x$, leave $a$ alone and replace $x$ by $c$. Then differentiate wrt $x$ etc.
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
$endgroup$
add a comment |
$begingroup$
Sometimes changing the notation makes it easier. Another way of verifying what you said is replace $b$ by $x$, leave $a$ alone and replace $x$ by $c$. Then differentiate wrt $x$ etc.
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
$endgroup$
Sometimes changing the notation makes it easier. Another way of verifying what you said is replace $b$ by $x$, leave $a$ alone and replace $x$ by $c$. Then differentiate wrt $x$ etc.
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
answered Jan 19 '15 at 14:53
AngeloAngelo
1796
1796
add a comment |
add a comment |
$begingroup$
hint: differentiating $lnleft(frac{a(b-x)}{b(a-x)}right)$ with respect to $b$ we get
$$frac{x}{b(b-x)}$$
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
$begingroup$
hint: differentiating $lnleft(frac{a(b-x)}{b(a-x)}right)$ with respect to $b$ we get
$$frac{x}{b(b-x)}$$
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
$begingroup$
hint: differentiating $lnleft(frac{a(b-x)}{b(a-x)}right)$ with respect to $b$ we get
$$frac{x}{b(b-x)}$$
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
hint: differentiating $lnleft(frac{a(b-x)}{b(a-x)}right)$ with respect to $b$ we get
$$frac{x}{b(b-x)}$$
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered Jan 19 '15 at 14:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
78.1k42867
add a comment |
add a comment |
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$begingroup$
It is correct. Good luck.
$endgroup$
– Galc127
Jan 19 '15 at 14:43