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So I'm not really sure whether I'm correct as several people are saying some of my syntax is wrong, where others are saying I have a wrong answer. I have checked my answer using Wolfram Alpha and it appears to be correct; could anyone please confirm/clarify?

Calculate the partial Derivative $\frac{\partial f}{\partial y}$ of $$f(x,y) = \frac{x}{y} \cos\left(\frac{1}{y}\right).$$

Is this correct? (Slightly reduced working because it's super long to type out)

$x$ is constant

$$\frac{\partial f}{\partial y} \left(\frac{\cos(\frac{1}{y})}{y}\right)$$

Quotient Rule

$$\frac{\dfrac{\partial f}{\partial y} \left(\dfrac{\cos(\frac{1}{y})}{y}\right)y-\dfrac{\partial f}{\partial y}(y) \cos(\frac{1}{y})}{y^2}$$

Chain Rule

$$\frac{\partial f}{\partial y} \left(\cos(\tfrac{1}{y})\right) = \frac{\sin(\frac{1}{y})}{y^2}$$

$$x\frac{\frac{\sin\frac{1}{y}}{y^2}y-1\cos(\frac{1}{y})}{y^2}$$

Simplified Answer

$$\frac{x \left(\sin\frac{1}{y} -y\cos(\frac{1}{y})\right)}{y^3}$$

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    $\begingroup$ You are wrong in the quotient rule. And he derivative of $\cos x$ is $-\sin x$, $\endgroup$ – Emilio Novati Jul 23 '15 at 20:56
  • $\begingroup$ Are you sure in this case as others have said this is correct? $\endgroup$ – Daniel Jul 23 '15 at 21:33
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    $\begingroup$ Making the substitution $z = \frac{1}{y}$ makes this a lot simpler $\endgroup$ – BadAtMaths Jul 23 '15 at 21:41
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    $\begingroup$ Yes. See my answer that is the same as Dr. Shonnard. $\endgroup$ – Emilio Novati Jul 23 '15 at 21:46
  • $\begingroup$ Hmm, I checked on two different online calculators my answer appears to be correct and I understand the step I have taken to get there; So I think ill use my answer. However the others are probably way quicker $\endgroup$ – Daniel Jul 23 '15 at 21:53
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$$\frac { \partial f\left( x,y \right) }{ \partial y } =\frac { \partial \left( \frac { x }{ y } \cos { \frac { 1 }{ y } } \right) }{ \partial y } =x\frac { \partial \left( \frac { 1 }{ y } \right) }{ \partial y } \cos { \frac { 1 }{ y } +\frac { x }{ y } \frac { \partial \left( \cos { \frac { 1 }{ y } } \right) }{ \partial y } } =\\ =-\frac { x }{ { y }^{ 2 } } \cos { \frac { 1 }{ y } +\frac { x }{ y^{ 3 } } \sin { \frac { 1 }{ y } } } =\frac { -x\left( y\cos { \frac { 1 }{ y } -\sin { \frac { 1 }{ y } } } \right) }{ { y }^{ 3 } } $$

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  • $\begingroup$ You have a wrong sign in the derivative of $\cos(1/x)$. $\endgroup$ – Emilio Novati Jul 23 '15 at 21:44
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    $\begingroup$ you mean cos(1/y),yes you are right thanks) $\endgroup$ – haqnatural Jul 23 '15 at 21:47
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by the chain and the sum rule we get $$\frac{x \sin \left(\frac{1}{y}\right)}{y^3}-\frac{x \cos \left(\frac{1}{y}\right)}{y^2}$$

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You have: $$ \dfrac{\partial}{\partial y} \dfrac{x \cos\left(\frac{1}{y}\right)}{y}=x \dfrac{\partial}{\partial y}\dfrac{ \cos\left(y^{-1}\right)}{y}=$$ $$ x\dfrac{-\sin y^{-1}(-y^{-2})y-\cos y^{-1}}{y^2}= $$ $$ x\dfrac{ \sin y^{-1}-y\cos y^{-1}}{y^3} $$

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Just one typo in Quotient Rule. The Quotient Rule should look like this:

$$\frac{\frac{\partial f}{\partial y}(cos(\frac{1}{y}))y-\frac{\partial f}{\partial y}(y) cos(\frac{1}{y})}{y^2}$$

Everything else is fine.

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  • $\begingroup$ Oh is the cos X should be - Sin X right? $\endgroup$ – Daniel Jul 23 '15 at 21:02
  • $\begingroup$ Just fix the first term of the nominator. $\endgroup$ – Ana Uspekova Jul 23 '15 at 21:21
  • $\begingroup$ Oh yes sorry, I just read other comments and wondered if thats what you meant... thanks $\endgroup$ – Daniel Jul 23 '15 at 21:21
  • $\begingroup$ Someone above said that the derivative of $\cos x$ is $-\sin x$ but you are correct because the derivative of $\frac{1}{y}$ is $-\frac{1}{y^2}$. $\endgroup$ – Ana Uspekova Jul 23 '15 at 21:45
  • $\begingroup$ I checked with wolfram alpha and my version seems to be correct. So I'm not sure whether Im correct or not $\endgroup$ – Daniel Jul 24 '15 at 17:25

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