Consider the function f(x,y)=x^2+y^2 where x=sin2θ and y=cos2θ

df/dθ is given by?(Headbang)

I know this involves the chain rule and trig id.

But I'm still not getting it

Kindly assist.

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- Jan 10th 2009, 06:49 AMKayPeePartial Differentiation
Consider the function f(x,y)=x^2+y^2 where x=sin2θ and y=cos2θ

df/dθ is given by?(Headbang)

I know this involves the chain rule and trig id.

But I'm still not getting it

Kindly assist. - Jan 10th 2009, 07:00 AMJester
Since $\displaystyle f$ is a function of $\displaystyle x \; \text{and}\; y$ and $\displaystyle x \; \text{and}\; y$ are functions of $\displaystyle \theta$, then $\displaystyle f$ is a function of $\displaystyle \theta$ so $\displaystyle \frac{df}{d \theta}$ makes sense. The chain rule is

$\displaystyle (1)\;\;\;\;\;\frac{df}{d \theta} = \frac{\partial f}{\partial x} \frac{d x}{d \theta} + \frac{\partial f}{\partial y} \frac{dy} {d \theta} $.

So calculate all of the derivatives

$\displaystyle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{d x}{ d \theta} \; \text{and}\;\frac{d y}{d \theta} $

and substitute them into (1). Finally, use your expressions for $\displaystyle x\; \text{and}\; y$ to get your final answer. - Jan 10th 2009, 07:10 AMProve It
- Jan 10th 2009, 08:55 AMKayPeeanswer
- Jan 10th 2009, 12:36 PMmr fantastic