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Determine f′(x) for f(x)=(ln(3x))^9.
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in other words find dy/dx for f(x)=(ln(3x))^9.
Do the chain rule. f(x)=g(h(x)) f'(x) = g'(h(x))*h'(x) if your f(x) = (ln(x^2))^2 your h(x)=ln(x^2) your g(x)= (h(x))^2 f'(x) in this case = \[2*(ln(x^2))*\frac{1}{ln(x^2)}*2x\] In this case, I had to do the chain rule twice. Notice the x^2 in the ln(x^2) had to be differentiated as well.
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