# Question Related to Partial Derivatives

If $$u=f(x,y)$$, where $$x=e^{2s}\cos(5t)$$ and $$y=e^{2s}\sin(5t)$$, then $$\Big(\frac{\partial u}{\partial x}\Big)^2+\Big(\frac{\partial u}{\partial y}\Big)^2=g(s,t)\Big(\frac{\partial u}{\partial s}\Big)^2+h(s,t)\Big(\frac{\partial u}{\partial t}\Big)^2$$. What are $$g(s,t)$$ and $$h(s,t)$$?

So I got that $$u=f\left(e^{2s}\cos(5t),\,e^{2s}\sin(5t)\right)$$ and that $$\frac{\partial u}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$$.

This gave me $$\frac{\partial u}{\partial s}=\frac{\partial f}{\partial x}\left(2e^{2s}cos\left(5t\right)\right)+\frac{\partial f}{\partial y}\left(2e^{2s}sin\left(5t\right)\right)$$.

Of course, I can also do the same thing with $$\frac{\partial u}{\partial t}$$, but I am not entirely sure on how to move on from this. How exactly will I go about squaring these partial derivatives and separating them to get $$g(s, t)$$ and $$h(s, t)$$? Or am I actually supposed to find the second partial derivative?

Any help or guidance would be highly appreciated!

The notation here is not very clear, so I have slightly changed it:

Let $$X$$ and $$Y$$ be the functions $$X(s,t)=e^{2s}\cos(5t)$$ and $$Y(s,t)=e^{2s}\sin(5t)$$, and $$u$$ be the function

$$u(s,t)=f[X(s,t),Y(s,t)]$$

What you have been asked to find is the $$g$$ and $$h$$ in:

$$f_x^2+f_y^2=g(s,t)u_s^2+h(s,t)u_t^2 \tag{1}$$ where I have suppressed the arguments of all the derivatives.

Here

$$u_s(s,t)=f_xX_s+f_yY_s\qquad \text{and}\qquad u_t(s,t)=f_xX_t+f_yY_t$$

where $$X_s(s,t)=2e^{2s}\cos(5t)=2X\qquad \text{and}\qquad Y_s(s,t)=2e^{2s}\sin(5t)=2Y$$ while $$X_t(s,t)=-5e^{2s}\sin(5t)=-5Y\qquad \text{and}\qquad Y_t(s,t)=5e^{2s}\cos(5t)=5X$$

So

$$u_s(s,t)=2(f_xX+f_yY)\qquad \text{and}\qquad u_t(s,t)=-5(f_xY-f_yX)$$

You can solve this system of equations to get

$$f_x=\frac{5Xu_s-2Yu_t}{10(X^2+Y^2)}\qquad \text{and}\qquad \frac{2Xu_t+5Yu_s}{10( Y^{2}+X^{2})}$$

Then

$$f_x^2+f_y^2=\frac{25u_s^2+4u_t^2}{100(X^2+Y^2)}=\frac{25u_s^2+4u_t^2}{100e^{4s}}$$

and so $$g(s,t)=\frac{1}{4e^{4s}}$$ and $$h(s,t)=\frac{1}{25e^{4s}}$$.

I suspect there is probably a quicker way...