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Complicated Derivative Problem:
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h(x) = f^3(g(x)) h'(x) = 3f^2(g(x)) * g'(x) by the chain rule
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h(x) = (f(g(x)))^3 h'(x) = 3(f(g(x)))^2 * f'(g(x)) * g'(x)
yup okay I got that far
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wait...oh I see i made an error when using the chain rule
h'(1) = 3(f(g(1)))^2 * f'(g(1)) * g'(1) h'(1) = 3(f(3))^2 * f'(3) * 3 h'(1) = 3(2)^2 * 5 * 3
h'(1) = 180 by chance?
thats strange, I got 300, let me check my work :O
isnt g'(1)=5?
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instead of 3?
yeah, trying to flip back and forth between windows tends to mess me up alot more than usual
lol XD sorry about that
concept is good tho :)
Thanks for the help!
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