# Thread: Finding a partial derivative using differentials

1. ## Finding a partial derivative using differentials

Hi,

I'll just get straight into the question then say what I'm confused about at the bottom...

x = rcos(t) and y = rsin(t), find dr/dx and dr/dy (partial derivatives)

The way to solve it in my notes goes:

dx = (dx/dr)dx + (dx/dt)dt = cos(t)dr - rsin(t)dt (sorry, don't know how to do partial d's)

cos(t)dx = cos^2(t)dr -rsin(t)cos(t)dt

dy = sin(t)dr + rcos(t)dt

sin(t)dy = sin^2(t)dr + rsin(t)cos(t)dt

so: cos(t)dx + sin(t)dy = dr
= (dr/dx)dx + (dr/dy)dy

-> (dr/dx) = cos(t) and (dr/dy) = sin(t)

I get all that, but if we go back to x = rcos(t) and y = rsin(t), couldn't we just rearrange these to make r the subject, the take the partial derivatives of that? But this gets (dr/dx) = 1/cos(t) and (dr/dy) = 1/sin(t), so it obviously can't be right, but where is it incorrect? I'm probably missing something simple...

Thanks

2. ## Re: Finding a partial derivative using differentials

x = rcos(t) and y = rsin(t), find dr/dx and dr/dy (partial derivatives)...........I get all that, but if we go back to x = rcos(t) and y = rsin(t), couldn't we just rearrange these to make r the subject, the take the partial derivatives of that? But this gets (dr/dx) = 1/cos(t) and (dr/dy) = 1/sin(t), so it obviously can't be right, but where is it incorrect? I'm probably missing something simple...
Partial derivative of what?

If you meant to take the derivative with respect to $\displaystyle x$ and $\displaystyle y$ of the function $\displaystyle r$ composed of $\displaystyle x$ and $\displaystyle y$ Then:

$\displaystyle x^2 + y^2 = r^2(\cos^2{(t)} + \sin^2{(t)}) = r^2......\text{[by pluggin' in the values of } x \text{ and } y \text{ and by trigonometry]}$

Then do implicit differentiation of $\displaystyle r^2 = x^2 + y^2$ with respect to $\displaystyle x$ which is:

$\displaystyle 2r\frac{d r}{d x} = 2x \,\,\,\therefore \frac{d r}{d x} = \frac{x}{r} = \cos{(t)}.....\text{[after rearrangement and substituting }\dfrac{x}{r}\text{ ]}$

And then with respect to $\displaystyle y$:

$\displaystyle 2r\frac{d r}{d y} = 2y \,\,\, \therefore \frac{d r}{d y} = \frac{y}{r} = \sin{(t)}......\text{[after rearrangement and substituting }\dfrac{y}{r}\text{ ]}$

Or if you meant $\displaystyle \frac{dr}{dx}$ of $\displaystyle x = r \cos{(t)}$ and $\displaystyle \frac{dr}{dy}$ of $\displaystyle y = r \sin{(t)}$ , then the answer $\displaystyle \frac{dr}{dx} = \frac{1}{\cos{(t)}}$ and $\displaystyle \frac{dr}{dy} = \frac{1}{\sin{(t)}}$ is correct.

Two different answers based on which function you choose to differentiate.