# Find the value of differentiability of function 18x^2+4

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

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?

• So you did $f'(6x) \cdot 6=18x^2+4$? If so, replace $x$ with $\frac{1}{3}$ and solve for $f'(2)$ – randomgirl Dec 6 '17 at 2:55
• Why 1/3? How did i get that? – nerv21 Dec 6 '17 at 2:58
• $f'(6x)=f'(2)$ when $x=\frac{1}{3}$ – randomgirl Dec 6 '17 at 2:58

$6f'(6x) = 18x^2 + 4$ (Using Chain Rule on the L.H.S.)
put $x = 1/3$ , we get,
$6f'(2) = 18/9 + 4$
$f'(2) = 1$