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anysportzchik
Lv 5
anysportzchik asked in Science & MathematicsMathematics · 1 decade ago

AP Calculus Math Problem?

Here is the given information

When x is = to 1, f(x)=3, f ' (x) = -2, g(x)=-3 g ' (x)= 4

If h(x)= (2f(x)+3)(1+g(x)) then what is h ' (1)

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  • Anonymous
    1 decade ago
    Favorite Answer

    Note that, By the Product Rule, if:

    y = a(x)*b(x).

    Then:

    dy/dx = a'(x)*b(x) + a(x)*b'(x).

    We see that:

    a(x) = 2f(x) + 3 ==> a'(x) = 2f'(x)

    b(x) = 1 + g(x) ==> b'(x) = g'(x).

    We now have:

    h'(x) = a'(x)*b(x) + a(x)*b'(x)

    ==> h'(x) = 2f'(x)*[1 + g(x)] + g'(x)*[2f(x) + 3].

    Finally, with the given information:

    h'(1) = 2f'(1)*[1 + g(1)] + g'(1)*[2f(1) + 3]

    ==> h'(1) = (-4)(1 - 3) + (4)[2(3) + 3] = 8 + 36 = 44.

    I hope this helps!

  • MechEng2030
    Lv 7
    1 decade ago

    h(x) = 2*f(x) + 2*f(x)g(x) + 3 + 3*g(x)

    h'(x) = 2*f'(x) + 2*[f(x)*g'(x) + g(x)*f'(x)] + 3*g'(x)

    h'(1) = 2(-2) + 2(12 + 6) + 3*(4)

    h'(1) = -4 + 36 + 12 = 44

    Source(s): Note to Brian: Isn't f'(x) -2? you made f'(x) equal -1 I think and that's how you got 40.
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