Implicit Differentiation Problem: x^2 + zsin(xyz) = 0

[hank]

Ok, here goes..

Instructions:
Calculate fx and fy using implicit differentiation. Leave answers in terms of x, y, z.

Problem:
x^2 + zsin(xyz) = 0

Comments:
I get close to finding the solution, but can't quite get my answer like the book's answer.
Can someone do this problem and show me the steps so I can see where I'm going wrong?

 

[galactus]

Re: Implicit Differentiation Problem

\(\displaystyle f_{x}\)o it the same as 'regular differentiation except regard the variables except for x as constants.

\(\displaystyle x^{2}+zsin(xyz)\)

\(\displaystyle f_{x}=2x+z^{2}ycos(xyz)\).........[1]

When you differentiate zsin(xyz), use the chain rule and remember that z is a constant. Derivative of outside is

\(\displaystyle cos(xyz)\). Since yz is a constant, per the product rule, the derivative of the inside is x(0)+zy.

So, we have \(\displaystyle z^{2}ycos(xyz)\). Add on the derivative of x^2 and you get [1]

 

[hank]

Re: Implicit Differentiation Problem

Hi, galactus..

I'm sorry, it was supposed to be done using implicit differentiation.
I put partial by mistake in the instructions.

 

[skeeter]

Re: Implicit Differentiation Problem

\(\displaystyle x^2 + z\sin(xyz) = 0\)

\(\displaystyle f_x = 2x + z\cos(xyz)\left[yz + x\frac{dy}{dx}z + xy\frac{dz}{dx}\right] + \sin(xyz)\frac{dz}{dx}\)

\(\displaystyle f_y = 2x \frac{dx}{dy} + z\cos(xyz)\left[\frac{dx}{dy}yz + xz + xy\frac{dz}{dy}\right] + \sin(xyz)\frac{dz}{dy}\)