0
$\begingroup$

The problem I found on the site was:

Let $h(x) = 4\sqrt[4]{x^5}$. Find $h'(x)$.

I got the answer $h'(x) = \frac{5x^4}{\sqrt[4]{x^{15}}}$ and the site told me it was the correct answer. However, when I looked at what the correct answer was it told me it was $5x^\frac{1}{4}$. Therefore, I would assume that $\frac{5x^4}{\sqrt[4]{x^{15}}} = 5x^\frac{1}{4}$, but I don't understand how? The only simplification I could determine is $\frac{5x^4}{x^{\frac{15}{4}}}$. Can somebody show me the step by step simplification to get from my answer to their answer?

|cite|improve this question
$\endgroup$
  • $\begingroup$ is this $h(x)=4\sqrt[4]{x^5}$? $\endgroup$ – Dr. Sonnhard Graubner Nov 13 '17 at 14:58
  • 2
    $\begingroup$ Since you're studying calculus then surely you must know that $\dfrac{x^a}{x^b} = x^{a-b}$ for nonzero $x$. What's $4 - \dfrac{15}4$? $\endgroup$ – tilper Nov 13 '17 at 14:59
  • $\begingroup$ @Dr. Sonnhard Graubner, yes $\endgroup$ – Ramzan Chasygov Nov 13 '17 at 15:00
  • $\begingroup$ @tilper, I didn't know that or must be forgot, thank you, now everything make perfectly sense) $\endgroup$ – Ramzan Chasygov Nov 13 '17 at 15:07
3
$\begingroup$

Note that $h(x) = 4x^{5/4}$ so that

$$ h'(x) = 4\cdot \frac{5}{4}\cdot x^{5/4-1} = 5x^{1/4} $$

EDIT: By using the chain rule:

$$ h'(x) = \frac{4}{4\sqrt[4]{x^{5\cdot 3}}}\frac{d (x^5)}{dx} = 5 \frac{x^4}{\sqrt[4]{x^{15}}} = 5 \frac{x^4}{x^{15/4}} = 5x^{4-15/4}=5x^{1/4} $$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I solve with chain rule and that why get more complex answer) $\endgroup$ – Ramzan Chasygov Nov 13 '17 at 15:13
  • $\begingroup$ @RamzanChasygov I updated my answer, hopefully it makes more sense now $\endgroup$ – caverac Nov 13 '17 at 15:29

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.