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'
f(x) = x^2*(2x+1)^3
(need chain rule, too)
In the form
y-y1 = m(x-x1)
y+1 = (10x^2+2x)(2x+1)^2(x+1)