Find h'(-3) if h= (f)(g), f'(x)= -10x^2-9, and the equation of the tangent line of g at -3 is y= -8x - 1

Answer 1

If h = fg , then h' = fg' + gf'

For h'(-3) you need f(-3) and f'(-3), g(-3) and g'(-3)

g is easier of the two: g'(-3) is the slope of the tangent line = -8 and g(-3) is y(-3) = 23

f'(-3) is also by subbing in -3 for x: = -99

I don't know if it was intentional, but you can only find f(x) to within a constant:

f(x) = int(-10x²-9) = -10x³/3 - 9x + C

f(-3) = -63 + C

Combine it all in the first equation.

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