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$\frac{d}{dx} f(6x)={18x^2 + 4}$

Find $f'(2)$

I don't know how to begin with this. First I took $f'(6x)*6$ and equalized it equal to the given equation. After that I took $u=6x$ and placed $2$ inside but that doesn't go with answers. Help pls?

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  • $\begingroup$ So you did $f'(6x) \cdot 6=18x^2+4$? If so, replace $x$ with $\frac{1}{3}$ and solve for $f'(2)$ $\endgroup$ – randomgirl Dec 6 '17 at 2:55
  • $\begingroup$ Why 1/3? How did i get that? $\endgroup$ – nerv21 Dec 6 '17 at 2:58
  • $\begingroup$ $f'(6x)=f'(2)$ when $x=\frac{1}{3}$ $\endgroup$ – randomgirl Dec 6 '17 at 2:58
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No substitutions are needed,

$ 6f'(6x) = 18x^2 + 4 $ (Using Chain Rule on the L.H.S.)

put $ x = 1/3 $ , we get,

$ 6f'(2) = 18/9 + 4 $

$ f'(2) = 1 $

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