# Chain Rule-Partial Derivatives!

• Jan 6th 2009, 03:57 PM
Pringles
Chain Rule-Partial Derivatives!
Consider the function f(x,y)=x^2+y^2 where x=sin2θ and y=cos2θ
df/dθ is given by?

Hello

I have studied stuff like this in 8 years.I've running around this for the past three days.

I have looked at the chain rule for multivariate functions.

The Trig identity Sin^2+Cos^2=1

I seem to be moving in circles
• Jan 6th 2009, 04:05 PM
Chris L T521
Quote:

Originally Posted by Pringles
Consider the function f(x,y)=x^2+y^2 where x=sin2θ and y=cos2θ
df/dθ is given by?

Hello

I have studied stuff like this in 8 years.I've running around this for the past three days.

I have looked at the chain rule for multivariate functions.

The Trig identity Sin^2+Cos^2=1

I seem to be moving in circles

By definition:

$\displaystyle \frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x}\cdot\frac{\,dx}{\,d\theta}+\frac{\partial f}{\partial y}\cdot\frac{\,dy}{\,d\theta}$

Since $\displaystyle f\left(x,y\right)=x^2+y^2$, when we partially differentiate the function wrt one variable, we hold the other variable constant. As a result, $\displaystyle \frac{\partial f}{\partial x}=2x$. I leave it for you to find $\displaystyle \frac{\partial f}{\partial y}$. You also need to find $\displaystyle \frac{\,dx}{\,d\theta}$ and $\displaystyle \frac{\,dy}{\,d\theta}$. You can easily do these since x and y are defined in terms of theta, and also because this is ordinary differentiation.

When you plug all those values into the equation, try to get your result in terms of one variable, $\displaystyle \theta$.

Can you take it from here?
• Jan 6th 2009, 04:17 PM
Pringles
I'll try
Quote:

Originally Posted by Chris L T521
By definition:

$\displaystyle \frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x}\cdot\frac{\,dx}{\,d\theta}+\frac{\partial f}{\partial y}\cdot\frac{\,dy}{\,d\theta}$

Since $\displaystyle f\left(x,y\right)=x^2+y^2$, when we partially differentiate the function wrt one variable, we hold the other variable constant. As a result, $\displaystyle \frac{\partial f}{\partial x}=2x$. I leave it for you to find $\displaystyle \frac{\partial f}{\partial y}$. You also need to find $\displaystyle \frac{\,dx}{\,d\theta}$ and $\displaystyle \frac{\,dy}{\,d\theta}$. You can easily do these since x and y are defined in terms of theta, and also because this is ordinary differentiation.

When you plug all those values into the equation, try to get your result in terms of one variable, $\displaystyle \theta$.

Can you take it from here?

I will give it a shot.Thanks