Help with this Partial Derivative?

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Yes that's right.

In that case, I get e^xy(1+y), however the correct answer is slightly different: e^xy(1+xy). Why is the x in the correct answer if we treated it like a constant in the process?

Reply 41

Notnek

Original post by Exceptional

In that case, I get e^xy(1+y), however the correct answer is slightly different: e^xy(1+xy). Why is the x in the correct answer if we treated it like a constant in the process?

What's the derivative of

e^{xy}

wrt y?

Reply 42

RDKGames

Study Forum Helper

Original post by Exceptional

In that case, I get e^xy(1+y), however the correct answer is slightly different: e^xy(1+xy). Why is the x in the correct answer if we treated it like a constant in the process?

\displaystyle \frac{\partial }{\partial y} e^{xy} \neq e^{xy}

Note that

\frac{d}{dx}e^{ax}=ae^{ax}

where

a=\mathrm{const.}

(edited 6 years ago)

Reply 43

Zacken

Original post by Exceptional

In that case, I get e^xy(1+y), however the correct answer is slightly different: e^xy(1+xy). Why is the x in the correct answer if we treated it like a constant in the process?

The product rule is

\displaystyle \frac{d}{dy}(y) e^{xy} + y\frac{d}{dy}(e^{xy}) = e^{xy} + yxe^{xy}

since

\frac{d}{dy} e^{2y} = 2e^{2y}

\frac{d}{dy}e^{xy} = xe^{xy}

Reply 44

Exceptional

Original post by Notnek

What's the derivative of

e^{xy}

wrt y?

xe^xy?

Reply 45

Notnek

Original post by Exceptional

xe^xy?

That's correct now. Now try the question again.

Reply 46

Exceptional

Original post by Notnek

That's correct now. Now try the question again.

Got it, makes sense now. Thanks!

Reply 47

MR1999

Original post by RDKGames

How about this, if you know about implicit differentiation.

z=\sqrt{x(y-2)} \Rightarrow z^2=x(y-2)

Taking the partial derivative wrt x of the LHS leaves you with

\displaystyle 2z\frac{\partial z}{\partial x}

Taking the partial derivative wrt x of the RHS leaves you with

y-2

So then

\displaystyle \frac{\partial z}{\partial x} = \frac{y-2}{2z} =\frac{1}{2}\cdot \frac{y-2}{\sqrt{x(y-2)}}=\frac{1}{2}\cdot \frac{y-2}{\sqrt{x}\sqrt{(y-2)}}

Thus you have

\displaystyle \frac{\partial z}{\partial x} =\frac{1}{2}\cdot \frac{\sqrt{y-2}}{\sqrt{x}}=\frac{1}{2}\sqrt{ \frac{y-2}{x} }

Get it?

This looks absolutely beautiful.

Reply 48

Exceptional

Original post by Notnek

That's correct now. Now try the question again.

Hi again, do you think you could help with this? My values for x = 1.25, y = .75, and lambda = .25. I rearranged and solved but must've gone wrong somewhere?