Use implicit differentiation to find dz/dx and dz/dy?

Question

Use implicit differentiation to find dz/dx and dz/dy for 3x^2*z+2z^3-3yz=0.

Answer: dz/dx=-6xz/(3x^2+6z^2-3y)

dz/dy=3z/(3x^2+6z^2-3y)

This is Calculus 3 topic. Show your work through steps.

Answer 1

Sun,

Hello :) Let's take a look at this problem.

3zx^2 + 2z^3 - 3yz = 0

Let's start by moving the 3yz over:

3zx^2 + 2z^3 = 3yz

Implicit differentiation is really just differentiating without making an explicit statement. In other words, we will leave this equation "as is" instead of restating to say that x = "this" or y = "that". Let's differentiate for dz/dx:

d/dx (3zx^2) + d/dx (2z^3) = d/dx (3yz)

3 d/dx (zx^2) + 2 d/dx (z^3) = 3 d/dx (yz)

Use product rule ( (uv)' = uv' + u'v ) for d/dx zx^2. For d/dx(yz), remember that when using implicit differentiation, if differentiating with respect to x (finding dz/dx), treat y as a constant.

3(z * 2x + dz/dx * x^2) + 2(3z^2)(dz/dx) = 3y * d/dx z

3( 2xz + x^2(dz/dx) ) + 6z^2(dz/dx) = 3y(dz/dx)

6xz + 3x^2(dz/dx) + 6z^2(dz/dx) = 3y(dz/dx) (Let's round up the dz/dx terms)

3x^2(dz/dx) + 6z^2(dz/dx) - 3y(dz/dz) = -6xz

dz/dx (3x^2 + 6z^2 - 3y) = -6xz

dz/dx = -6xz / (3x^2 + 6z^2 - 3y)

As was to be shown. The process is the same of course for determining dz/dy. Again, remember that since we are now differentiating with respect to y, we will treat x as a constant.

Find dz/dy for 3zx^2 + 2z^3 - 3yz = 0

3zx^2 + 2z^3 = 3yz

d/dy 3zx^2 + d/dy 2z^3 = d/dy 3yz

3 d/dy zx^2 + 2 d/dy z^3 = 3 d/dy yz (Now treat x as a constant when differentiating, use product rule for d/dy yz)

3x^2 d/dy z + 2(3z^2)(dz/dy) = 3(y(dz/dy) + 1(z))

3x^2(dz/dy) + 6z^2(dz/dy) = 3y(dz/dy) + 3z (Round up dz/dy)

3x^2(dz/dy) + 6z^2(dz/dy) - 3y(dz/dy) = 3z

dz/dy ( 3x^2 + 6z^2 - 3y ) = 3z

dz/dy = 3z / ( 3x^2 + 6z^2 - 3y )

As was to be shown! Hope this helps :)

-Skyler