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AP CALCULUS

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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 - ,

    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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