AP CALCULUS
posted by Naz on .
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).

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
yy1 = m(xx1)
y+1 = (10x^2+2x)(2x+1)^2(x+1)