# What does $f(5x)$ mean when I differentiate?

if $$f(5x)=x^2 + x$$

$$f '(2) = ?$$

But I do not know how it came

How i can factor or simplify $$f(5x)$$ but i want understand what f(5x) mean?

• chain rule :$5f'(5x)=2x+1$ put $x=2/5$. here – daulomb Dec 9 '17 at 11:54
• Firstly, you need $f(y) = y^2/25 + y/5$. – Ryan Dec 9 '17 at 11:55
• that's wrong the answer is 9/25 when x=2 i want know how ! – john wick Dec 9 '17 at 12:04
• I think the point here is finding $f'(2)$ without finding a general formula for $f(y)$. – GEdgar Dec 9 '17 at 12:07
• @johnwick But we just left you $2$ answers below. – Rebellos Dec 9 '17 at 12:08

The derivative of the expression $f(g(x))$ is given as :

$$[f(g(x))]'=f'(g(x))\cdot g(x)$$

supposing that $f,g$ are differentiable functions.

This means that :

$$[f(5x)]'=f'(5x)\cdot (5x)'=5f'(5x)$$

which means, taking the derivative on both sides of $f(5x) = x^2 + x$, you'll get :

$$[f(5x)]'=\frac{2}{5}x+\frac{1}{5}$$

Now, just setting $x=2/5$, you'll yield your result for $f'(2) = 9/25$.

This is important to know (the chain rule) because such differentation issues occur often.

Alternatively, you could just let $t=5x$ and substitute and calculate from there on.

• Thanks for your answer. [+1] Yet, when setting $x = 2/5$, you still assume there is a $y = 5x$ in your mind. – Ryan Dec 9 '17 at 11:59
• @Ryan Sure I do, yes. Better not to say that $y=5x$ by the way, since $y=f(x)$ in the cartesian coordinates system. Let's say that in order to get the number $2$ from the expression $5x$ you'd need $x=2/5$. Alternatively, it just follows from letting $t=5x$ and substituting. – Rebellos Dec 9 '17 at 12:07
• Yes, you are right. Using $y$ is indeed confusing. Thanks :-) – Ryan Dec 9 '17 at 12:11
• @johnwick I have edited Rebellos' answer, and please wait it for peer review. – Ryan Dec 9 '17 at 12:28
• @GNUSupporter No worries :-) – Ryan Dec 9 '17 at 13:00

Setting $$t=5x$$ then we get $$f(t)=\left(\frac{t}{5}\right)^2+\frac{t}{5}$$ then we get $$f'(t)=...$$

• yes Dr this is the correct answer thx but why we setting t = 5x ? – john wick Dec 9 '17 at 12:36
• @johnwick This is probably because you need $5x = 2$ and $2$ can be regarded as $t = 2$. – Ryan Dec 9 '17 at 12:40
• @Ryan but i want (5x)' =2 Is there a difference – john wick Dec 9 '17 at 12:50
• @johnwick $(5x)' = 5$ and it cannot be $2$. – Ryan Dec 9 '17 at 12:52
• @johnwick You do not want $5x=2$, you want $f'(5x) \to f'(2)$ which means $5x=2 \Leftrightarrow x = 2/5$ – Rebellos Dec 9 '17 at 13:13