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

User Feraz
by
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1 Answer

3 votes

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))

User Manuella
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