Posted by **Naz** on Sunday, November 7, 2010 at 12:38pm.

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).

- AP CALCULUS -
**EmoryU**, Sunday, November 7, 2010 at 10:17pm
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)

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