Partial derivatives problem

**john1991** · April 21st 2010, 02:05 PM

1. a) Calculate the partial derivatives for $f(x,y)=x^2 siny$ and prove that

b)If $x(u)=u^2$ and $y(u)=1/u$ , calculate $df/du$ using the chain rule.

c)For $g(r)=r^4$ where $r=sqrt(x^2+y^2+z^2)$ use the chain rule to calculate $dg/dx$ , $d^2g/dx2$ and $d^2g/dxdy$ .

I am having a lot of trouble with this problem.

**undefined** · April 21st 2010, 02:58 PM

Originally Posted by john1991

1. a) Calculate the partial derivatives for $f(x,y)=x^2 siny$ and prove that

b)If $x(u)=u^2$ and $y(u)=1/u$ , calculate $df/du$ using the chain rule.

c)For $g(r)=r^4$ where $r=sqrt(x^2+y^2+z^2)$ use the chain rule to calculate $dg/dx$ , $d^2g/dx2$ and $d^2g/dxdy$ .

I am having a lot of trouble with this problem.

a) To find the partial derivative of f(x,y) with respect to x, treat y as a constant, and vice versa. Have you tried this?

b) Maybe someone else can help me out here because I don't see how the chain rule applies. Replace all instances of x with x(u) = u^2 and likewise for y in f(x,y). You are left with a univariate problem of finding the derivative of f(u) with respect to u. (Here we are overloading the variable name f as in: f(u) = f(x(u),y(u)), which may be confusing, but that seems to be the notation of the problem.)

c) First of all, g(r) simplifies to $g(x,y,z)=(x^2+y^2+z^2)^2$ . For the first partial derivative, you'll solve exactly as you would

d/dx (x^2 + c)^2

As in,

d/dx (x^2 + c)^2 = 2*(x^2 + c) * 2x

Hopefully it makes more sense now?

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