Finding g'(x) in a(x)=x(18-x^2)^1/2

[Shambles]

Homework Statement

Find g'(x)
a(x)=x(18-x^2)^1/2

Homework Equations

Answer stated as:
a'(x)=(18-x^2)^1/2 - x^2/(18-x^2)^1/2

The Attempt at a Solution

Having trouble with this solution. The chain rule states that f(x)g(x) = f'(x)g(x)+g'(x)f(x) so the first term in the solution is obviously (18-x^2)^1/2.

Where I run into problems is the second term with the numerator being x^2 when I thought it should only be x. Simplifying the problem into f(x)=(18-x^2)^1/2 the chain rule states it the derivative should be 1/2(18-x^2)(0-2x)=-x/(18-x^2)^1/2. Am I wrong or is the stated solution wrong?

 

[Doc Al]

Having trouble with this solution. The chain rule states that f(x)g(x) = f'(x)g(x)+g'(x)f(x) so the first term in the solution is obviously (18-x^2)^1/2.

Where I run into problems is the second term with the numerator being x^2 when I thought it should only be x.

You forgot about f(x) = x, which gives you the second x.

 

[Shambles]

Ugh, that is so... dirty. Thanks, that was pretty simple.

 

[Mark44]

Homework Statement

Find g'(x)
a(x)=x(18-x^2)^1/2

Homework Equations

Answer stated as:
a'(x)=(18-x^2)^1/2 - x^2/(18-x^2)^1/2

The Attempt at a Solution

Having trouble with this solution. The chain rule states that f(x)g(x) = f'(x)g(x)+g'(x)f(x) so the first term in the solution is obviously (18-x^2)^1/2.

That's not the chain rule; it's the product rule. The chain rule is used to find the derivative of a composite function, f(g(x)).

Where I run into problems is the second term with the numerator being x^2 when I thought it should only be x. Simplifying the problem into f(x)=(18-x^2)^1/2 the chain rule states it the derivative should be 1/2(18-x^2)(0-2x)=-x/(18-x^2)^1/2. Am I wrong or is the stated solution wrong?