Partial differentiation problem

1. Aug 12, 2015

DevonZA

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. If z=x+sin($x^2$y) + ln y find $\frac{\partial ^2z}{\partial x^2}$ and $\frac{\partial ^2z}{\partial y^2}$

2. Second order partial differentiation.

3. $\frac {\partial z}{\partial x}$ = 1 + $cos(x^2y)$ . (2x)
= $\frac{\partial ^2z}{\partial x^2}$ = $\frac {\partial }{\partial x}$ (1 + $cos(x^2y)$ . 2x)
= $-sinx^2$y.2x.2
= $-4xsinx^2$y

$\frac{\partial z}{\partial 2}$ = $cosx^2$ + $\frac{1}{y}$
$\frac{\partial ^2z}{\partial y^2}$ = $\frac{\partial }{\partial 2}$ ($cosx^2$ + $\frac{1}{y}$)
= $\frac{1}{y}$ or $y^{-1}$

2. Aug 12, 2015

SteamKing

Staff Emeritus
Is this correct? what is the derivative of x2 ?

3. Aug 12, 2015

SteamKing

Staff Emeritus
Is $\frac{\partial cos(x^2y)}{\partial y}=cos(x^2)$ ?

that would mean that $\frac{d cos(θ)}{dθ} = -sin(1)$

You should review the rules of differentiation and know them thoroughly before doing partials.

4. Aug 13, 2015

DevonZA

= $\frac{\partial ^2z}{\partial x^2}$ = $\frac {\partial }{\partial x}$ (1 + $cos(x^2y)$ . 2x)
= -2sinxy.2x.2
= -8xsinxy
Is that correct?

Partial differentiation is new to me..clearly

5. Aug 13, 2015

D H

Staff Emeritus
You presumably wouldn't have a problem were you asked to compute the ordinary derivatives $\frac{d^2}{dx^2}\left(x+\sin(x^2 b)+\ln b\right)$ and $\frac{d^2}{dy^2}\left(a + \sin(a^2 y) + \ln y\right)$, where $a$ and $b$ are some constants. Now replace $a$ and $b$ with $x$ and $y$ and voila! you have your partial derivatives. When computing the partial derivative of some function with respect to one of the independent variables, you simply treat the other independent variables as if they were constants.

6. Aug 13, 2015

SteamKing

Staff Emeritus
Yes, but I'm afraid the problem goes much deeper, back to taking ordinary derivatives.

What would be the derivative of this expression (no partials): $\frac{d}{dx}cos(x^2y)$ ?

7. Aug 13, 2015

DevonZA

$\frac{d}{dx}cos(x^2y)$ = -sin2x

8. Aug 13, 2015

RUber

You need to use the chain rule.
$\frac{d}{dx} f(g(x)) = f'(g(x) ) g'(x)$
So
$\frac{d}{dx} cos(x^2) = -sin(x^2) \frac{d}{dx} x^2$
Similarly,
$\frac{d}{dx} cos(x^2y) = -sin(x^2y) \frac{d}{dx} x^2y$

9. Aug 13, 2015

DevonZA

Thanks again RUber that makes sense. I'm out at dinner currently but I'll attempt to answer the question again when I'm home later.

10. Aug 13, 2015

DevonZA

Okay I have no idea if I am doing this right but let me know..

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11. Aug 13, 2015

SteamKing

Staff Emeritus
You can't have "sin" or "cos" standing by themselves. These are functions which must show an argument of some kind.

You've calculated $\frac{\partial z}{\partial x}$ correctly, but from that point on, it seems you are having trouble applying the chain rule correctly again.

Again, I repeat, you should not be working partial derivative problems unless and until you understand derivatives of a single variable thoroughly.

12. Aug 13, 2015

RUber

$\frac{d}{dx} x^2 y$ implies that you treat y as a constant. You can't just drop it out. It isn't like taking the derivative of x^2 +y.
Take your time. Be careful with your chain rule. You can get it done.

13. Aug 13, 2015

DevonZA

I had a feeling that the trig functions were wrong..
I struggle with derivatives so yes I need more practice but time is not on my side. I'll try again

14. Aug 13, 2015

DevonZA

So I haven't calculated $\frac {\partial{z}}{\partial{x}}$ correctly then?

$\frac{d}{dx} x^2y = 2xy$ ?

15. Aug 13, 2015

RUber

That's right.

16. Aug 14, 2015

DevonZA

Please see the attachment. If you could help me with the next step from here please, the trig functions are confusing me.

Attached Files:

• 2.1.pdf
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17. Aug 14, 2015

RUber

That looks like you might need to use the product rule. Then for the part of the product that has d/dx cos(x^2y), use the chain rule again.
You have already shown that you can use these methods. Take your time and do them one by one.

18. Aug 16, 2015

DevonZA

Final answer attached. Thanks to all who helped.