Question

How do you find the derivative of (x+1) times the third root of (x-5)spuared(x=4)?

Answer 1

Do you mean
f9x) = (x+1)[(x-5)^2*(x-4)]^1/3 ?
Which terms are included in the cube root?
The x=4 in parentheses makes no sense

Answer 2

Yes the eqaution you wrote is correct. There is no f9x however.

Answer 3

I meant f(x). It represent the "function of x" that you wrote.

Rewrite your function as
(x+1)*(x-5)^(2/3)*(x-4)^1/3

Use the rule that
d/dx (f*g*h) = f d/dx(g*h)+ gh df/dx
= fg dh/dx + fh dg/dx + gh df/dx
where f, g and h are functions of x.

That makes the derivative
(x-5)^(2/3)*(x-4)^1/3
+ (2/3)(x-4)(-1/3)*(x-1)(x-4)^(1/3)
+ (x+1)*(x-5)^(2/3)*(1/3)(x-4)^(-2/3)

Answer 4

Hey thanks a lot.