# Partial differentiation problem, multiple variables (chain rule?)

## Homework Statement

if z = x2 + 2y2 , x = r cos θ , y = r sin θ , find the partial derivative

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$

z = x2 + 2y2
x = r cos θ
y = r sin θ

## The Attempt at a Solution

The textbook says that the equation should be re-written to include only the variables θ and x, and then differentiated with respect to θ.

Substituting y = r sin θ :

z = x2 + 2r2 sin2 θ

then $\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = 4r2sin θ cos θ

However the solutions in the book give

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = 4r2 tan θ

What am I missing here?

pasmith
Homework Helper

## Homework Statement

if z = x2 + 2y2 , x = r cos θ , y = r sin θ , find the partial derivative

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$

z = x2 + 2y2
x = r cos θ
y = r sin θ

## The Attempt at a Solution

The textbook says that the equation should be re-written to include only the variables θ and x, and then differentiated with respect to θ.

Substituting y = r sin θ :

z = x2 + 2r2 sin2 θ

then $\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = 4r2sin θ cos θ

However the solutions in the book give

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = 4r2 tan θ

What am I missing here?

You need to differentiate $y^2 = r^2 \sin^2 \theta$ using the product rule: $r$ is not independent of $\theta$, since $$r = \frac{x}{\cos \theta}$$

Last edited:
• 1 person
Thanks! I used the product rule to differentiate, but I still think I'm missing something. My working is as follows:

z = x2 + 2r2 sin2 θ

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = $\frac{\partial}{\partial \theta}2r^{2}sin^{2}\theta$

$= 4r\frac{\partial r}{\partial \theta}sin^{2}\theta + 2r^{2}2sin\theta cos\theta$ (chain rule for dr/dθ)

since $r = \frac{x}{cos \theta}$, $\frac{\partial r}{\partial \theta} = \frac{-x sin \theta}{cos^{2} \theta}$

giving $-4r \frac{x sin \theta}{cos^{2} \theta} sin^{2} \theta + 4r^{2} sin \theta cos \theta$

I can replace x with r / cos θ, but I don't see how it reduces to the given solution of 4r2 tan θ.

Thanks again

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

if z = x2 + 2y2 , x = r cos θ , y = r sin θ , find the partial derivative

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$

z = x2 + 2y2
x = r cos θ
y = r sin θ

## The Attempt at a Solution

The textbook says that the equation should be re-written to include only the variables θ and x, and then differentiated with respect to θ.

Substituting y = r sin θ :

z = x2 + 2r2 sin2 θ

then $\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = 4r2sin θ cos θ

However the solutions in the book give

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = 4r2 tan θ

What am I missing here?

You are missing the fact that ##x## is held constant. One way to do it is:
$$dz = 2 x\, dx + 4 y \,dy\\ dx = \cos(\theta) \, dr - r \sin(\theta)\, d \theta\\ dy = \sin(\theta)\, dr + r \cos(\theta)\, d \theta$$
But ##dx = 0 \Longrightarrow dr = r \tan(\theta) \, d \theta##, so
$$dy = r \sin(\theta)\tan(\theta)\, d \theta + r \cos(\theta) \,d \theta = r \left( \frac{\sin^2(\theta)}{\cos(\theta)} + \cos(\theta)\right)\, d \theta = \frac{r}{\cos(\theta)}\, d \theta$$
Thus
$$dz = 4 y dy = 4 r \sin(\theta) (r/\cos(\theta)) \,d \theta = 4 r^2 \tan(\theta) \, d \theta$$
The partial ##(\partial z/\partial \theta)_{x} ## is the coefficient of ##d \theta## in the above.

• 1 person
pasmith
Homework Helper
Thanks! I used the product rule to differentiate, but I still think I'm missing something. My working is as follows:

z = x2 + 2r2 sin2 θ

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = $\frac{\partial}{\partial \theta}2r^{2}sin^{2}\theta$

$= 4r\frac{\partial r}{\partial \theta}sin^{2}\theta + 2r^{2}2sin\theta cos\theta$ (chain rule for dr/dθ)

since $r = \frac{x}{cos \theta}$, $\frac{\partial r}{\partial \theta} = \frac{-x sin \theta}{cos^{2} \theta}$

The derivative of $u^{-1}$ with respect to $u$ is $-u^{-2}$. The derivative of $\cos \theta$ with respect to $\theta$ is $-\sin \theta$. The two minus signs cancel.

I can replace x with r / cos θ

You can't, but you can replace $x$ with $r \cos \theta$ and do some trigonometric simplifications; the first step is to express everything in terms of sines and cosines.

• 1 person
Thanks! I used the product rule to differentiate, but I still think I'm missing something. My working is as follows:

z = x2 + 2r2 sin2 θ

$\left(\frac{\partial z}{\partial \theta}\right)_{x}$ = $\frac{\partial}{\partial \theta}2r^{2}sin^{2}\theta$

$= 4r\frac{\partial r}{\partial \theta}sin^{2}\theta + 2r^{2}2sin\theta cos\theta$ (chain rule for dr/dθ)

since $r = \frac{x}{cos \theta}$, $\frac{\partial r}{\partial \theta} = \frac{-x sin \theta}{cos^{2} \theta}$

giving $-4r \frac{x sin \theta}{cos^{2} \theta} sin^{2} \theta + 4r^{2} sin \theta cos \theta$

I can replace x with r / cos θ, but I don't see how it reduces to the given solution of 4r2 tan θ.

Thanks again

Try saying z = x2 + 2$\frac{x^2}{cos^2θ}$sin2θ = x2(1 + 2tan2θ)

Now with this for z you can perform $\left(\frac{\partial z}{\partial \theta}\right)_{x}$ quite easily.

Hint* Remember, that after you perform the derivation to look for anywhere you can make a substitution to remove 'x'.

• 1 person
The derivative of $u^{-1}$ with respect to $u$ is $-u^{-2}$. The derivative of $\cos \theta$ with respect to $\theta$ is $-\sin \theta$. The two minus signs cancel.

You can't, but you can replace $x$ with $r \cos \theta$ and do some trigonometric simplifications; the first step is to express everything in terms of sines and cosines.

That was a typo, my bad. So after sorting the minus sign, I'm left with essentially what I had before, but I can't see anyway of reducing $4r^{2} tan \theta sin^{2} \theta + 4r^{2} sin \theta cos \theta$ to $4r^{2} tan \theta$ without ending up with a huge mess.

Sorry for being dense :P

Figured it out. Thanks guys. Turns out I laid it out the way jaytech said, but didn't use the product rule properly so I abandoned that method and tried it another way, which lead to that whole mess. Whoops!

Thanks to everyone who helped!

You should try the steps I previously mentioned. Then reflect on why it works..

pasmith
Homework Helper
That was a typo, my bad. So after sorting the minus sign, I'm left with essentially what I had before, but I can't see anyway of reducing $4r^{2} tan \theta sin^{2} \theta + 4r^{2} sin \theta cos \theta$ to $4r^{2} tan \theta$ without ending up with a huge mess.

Sorry for being dense :P

$$\sin \theta \cos \theta = \tan\theta \cos^2 \theta$$

• 1 person
I did use the steps you suggested, jaytech. As for why it works, I imagine that I should be able to reach the solution from any starting point, with proper application of the chain rule/product rule, but arranging it as you suggested means I can skip over a lengthy simplification process after the operation. Any other insight you care to offer? :)