if $F'(ax) = G(x)$, then $F'(nax) = nG(nx)$?

Question

A question I found stated that:

Given $$\frac{d}{dx}F(ax) = G(x)$$

then $$\frac{d}{dx}F(2ax) = \dots$$

with choices:

          1.) G(2ax)
          2.) G(ax)
          3.) 2G(2x)
          4.) 2G(x)

I got the third option as the answer by assuming that

$F(ax) = \sin(ax)$ and therefore $F'(ax)= G(x) = a\cos(ax)$

and hence:

$F(2ax) = \sin(2ax)$ and therefore $F'(2ax)= 2a\cos(2ax) = 2G(2x)$

I then tested with $F(ax) = (ax)^2$ and $F(ax) = \log(ax)$ with same result as above.

Is there a general rule to proof the statement: if $F'(ax) = G(x)$, then $F'(nax) = nG(nx)$?

You are confusing $F'(bx)$ ($F$'s derivative evaluated at $ax$) and $(F(bx))'= b F'(bx)$ (the derivative of $F(bx)$) — reuns, Dec 18 '20 at 16:34

score 1 · Accepted Answer · answered Dec 19 '20 at 14:24

I repost my answer. In the last one I made a silly mistake.

Using the chain rule:$$\frac{d}{dx}F(2ax) =(2ax)' F'(2ax) = 2a F'(2ax)$$

Since we have for hypothesis that $$G(x) := \frac{d}{dx}F(ax) = aF'(ax) \quad \Rightarrow \quad G(2x) = \frac{d}{d(2x)}F(2ax) \stackrel{\star}{=} aF'(2ax)$$

If you do not understand the derivation $\star$ define $y := 2x$, then $$\frac{d}{d(2x)}F(2ax) = \frac{d}{d(y)}F(ay) = aF'(ay) = aF'(2ax)$$

You can now deduce from the equations above that we have $$\frac{d}{dx}F(2ax) = 2a F'(2ax) = 2 G(2x)$$

As @reuns pointed out in the comments you have to undarstand that there is a difference between $(F(ax))'$ and $F'(ax)$

if $F'(ax) = G(x)$, then $F'(nax) = nG(nx)$?

1 Answers1