2
$\begingroup$

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

$$f '(2) = ?$$

Answer is 9/25

But I do not know how it came

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

$\endgroup$
  • 1
    $\begingroup$ chain rule :$5f'(5x)=2x+1$ put $x=2/5$. here $\endgroup$ – daulomb Dec 9 '17 at 11:54
  • $\begingroup$ Firstly, you need $f(y) = y^2/25 + y/5$. $\endgroup$ – Ryan Dec 9 '17 at 11:55
  • $\begingroup$ that's wrong the answer is 9/25 when x=2 i want know how ! $\endgroup$ – john wick Dec 9 '17 at 12:04
  • $\begingroup$ I think the point here is finding $f'(2)$ without finding a general formula for $f(y)$. $\endgroup$ – GEdgar Dec 9 '17 at 12:07
  • $\begingroup$ @johnwick But we just left you $2$ answers below. $\endgroup$ – Rebellos Dec 9 '17 at 12:08
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for your answer. [+1] Yet, when setting $x = 2/5$, you still assume there is a $y = 5x$ in your mind. $\endgroup$ – Ryan Dec 9 '17 at 11:59
  • $\begingroup$ @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. $\endgroup$ – Rebellos Dec 9 '17 at 12:07
  • $\begingroup$ Yes, you are right. Using $y$ is indeed confusing. Thanks :-) $\endgroup$ – Ryan Dec 9 '17 at 12:11
  • $\begingroup$ @johnwick I have edited Rebellos' answer, and please wait it for peer review. $\endgroup$ – Ryan Dec 9 '17 at 12:28
  • 1
    $\begingroup$ @GNUSupporter No worries :-) $\endgroup$ – Ryan Dec 9 '17 at 13:00
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.