How do you differentiate f(x)=ln(sqrt(sin(x^2-3x)))f(x)=ln(√sin(x2−3x)) using the chain rule?

1 Answer
Bdub
Mar 4, 2016

f'(x)=1/(sqrt(sin(x^2-3x))) *1/(2sqrt(sin(x^2-3x))) cos (x^2-3x)(2x-3)
f'(x)=((2x-3)cos(x^2-3x))/(2sin(x^2-3x)) = 1/2 (2x-3)cot(x^2-3x)

Explanation:

You have four functions in here. Start with the ln and keep everything else the same then move to the sqrt, then sine and then x^2-3x. Then simplify note that cos(x^2-3x)/(sin(x^2-3x))=cot(x^2-3x)

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