Partial Differentiation

**KayPee** · January 10th 2009, 06:49 AM

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

I know this involves the chain rule and trig id.

But I'm still not getting it

Kindly assist.

**Danny** · January 10th 2009, 07:00 AM

Originally Posted by KayPee

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

I know this involves the chain rule and trig id.

But I'm still not getting it

Kindly assist.

Since $f$ is a function of $x \; \text{and}\; y$ and $x \; \text{and}\; y$ are functions of $\theta$ , then $f$ is a function of $\theta$ so $\frac{df}{d \theta}$ makes sense. The chain rule is

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

So calculate all of the derivatives

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

and substitute them into (1). Finally, use your expressions for $x\; \text{and}\; y$ to get your final answer.

**Prove It** · January 10th 2009, 07:10 AM

Originally Posted by KayPee

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

I know this involves the chain rule and trig id.

But I'm still not getting it

Kindly assist.

What does $\sin^2{2\theta} + \cos^2{2\theta}$ equal?

What is the derivative of a constant?

**KayPee** · January 10th 2009, 08:55 AM

Originally Posted by Prove It

What does $\sin^2{2\theta} + \cos^2{2\theta}$ equal? one

What is the derivative of a constant?

0

how would these two answers help me arrive at the final answer

thanks

**mr fantastic** · January 10th 2009, 12:36 PM

Originally Posted by KayPee

0

how would these two answers help me arrive at the final answer

thanks

Look at your expression for f(x, y). Now look at what Prove It wrote. Join the dots.

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