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Let f and g be differentiable functions such that f(1) = 4, g(1) = 3, f'(3) = -5, f'(1) = -4, g'(1) = -3, g'(3) = 2 If H(x) = f(g(x)), then h'(1) =
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Remember the chain rule. L(x)=f(g(x)) L'(x)=f'(g(x))g'(x)
take the derivative of f(g(x)). just treat them like they are variables. so you get: h'=f'(g(x))g'(x) now plug in your x value and evaluate: h'(1)=f'(g(1))(g'(1)) substitute in values that you know and evaluate again h'(1)=f'(3)(-3) h'(1)=(-5)(-3)=15
Suppose f and g are differentiable functions and that (f g) (x)= x^2/x-1, f(4)=2 and f'(4)=3/4 Determine the following a) g(4) and b) g'(4)
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