Answer:
[tex]y'=\frac{5}{x} \cdot \cos(\ln(2x^5))[/tex]
Step-by-step explanation:
[tex]y=\sin(\ln(2x^5))[/tex]
We are going to use chain rule.
The most inside function is [tex]y=2x^5[/tex] which gives us [tex]y'=10x^4[/tex].
The next inside function going out is [tex]y=\ln(x)[/tex] which gives us [tex]y'=\frac{1}{x}[/tex].
The most outside function is [tex]y=\sin(x)[/tex] which gives us [tex]y'=\cos(x)[/tex].
[tex]y'=10x^4 \cdot \frac{1}{2x^5} \cdot \cos(\ln(2x^5))[/tex]
[tex]y'=\frac{5}{x} \cdot \cos(\ln(2x^5))[/tex]
