Question

What is an equation of the line tangent to the graph of f(x)=x2(2x+1)3 at the point where x=-1?

(that's x squared (2x+1)cubed)

I'm having trouble finding the derivative:f'(x)which would be the slope (m).

Answer 1

To find the derivative, you must use the product rule
f'(x) = u'*v + u*v'

So,
f(x) = x^2*(2x+1)^3
f'(x)=(2x)(2x+1)^3+(x^2)(3)(2x+1)^2(2)
(need chain rule, too)

Simplify
(2x)(2x+1)^2(5x+1)
(10x^2+2x)(2x+1)^2

In the form
y-y1 = m(x-x1)
y+1 = (10x^2+2x)(2x+1)^2(x+1)