Y=sin(in(2x^5)) find the derivative

Question

asked Nov 7, 2020 204k views

1 Answer

Manuella · Answer 1 · 2020-11-12T02:45:14+0000

Answer:

$y'=(5)/(x) \cdot \cos(\ln(2x^5))$

Explanation:

$y=\sin(\ln(2x^5))$

We are going to use chain rule.

The most inside function is
y=2x^5 which gives us
y'=10x^4 .

The next inside function going out is
$y=\ln(x)$ which gives us
y'=(1)/(x) .

The most outside function is
$y=\sin(x)$ which gives us
$y'=\cos(x)$ .

$y'=10x^4 \cdot (1)/(2x^5) \cdot \cos(\ln(2x^5))$

$y'=(5)/(x) \cdot \cos(\ln(2x^5))$