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?

• is this $h(x)=4\sqrt[4]{x^5}$? Nov 13, 2017 at 14:58
• Since you're studying calculus then surely you must know that $\dfrac{x^a}{x^b} = x^{a-b}$ for nonzero $x$. What's $4 - \dfrac{15}4$?
– user307169
Nov 13, 2017 at 14:59
• @Dr. Sonnhard Graubner, yes Nov 13, 2017 at 15:00
• @tilper, I didn't know that or must be forgot, thank you, now everything make perfectly sense) Nov 13, 2017 at 15:07

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