Question

determine de value of dy/dx for the given value
a) y=(x^3)(3x+7)^2, x=-2
b) y= (2x+1)^5(3x+2)^4, x=-1

i know how to get the derivate but the exponents outside the brackets confuse me

Answer 1

From what you are saying about the exponents confusing you,
you clearly do not know how to get the derivatives.

did you get ...
y = x^3(3x+7)^2
dy/dx = x^3(2)(3x+7)(3) + 3x^2(3x+7)^2 , as your first-line derivative
using a combination of product rule and chain rule ???

This can of course be simplified, but since you just want the value
of dy/dx when x = -2, let's just sub that in

when x = -2
dy/dx = (-2)^3 (2)(-6+7)(3) + 3(4)(-6+7)^2
= -48 + 12
= -36

state what your derivative is for the 2nd equation.

Answer 2

then i dont know thats why im studying it :) how did you get the derivate of (3x+7)^2?

Answer 3

the chain rule says that if y = u^n
where u is a function of x, then
dy/dx = n u^(n-1) du/dx
If you look carefully at the calculated derivative, that is what mathhelper has done.