Calculus · real student question

Differentiate y = log(sin x) with respect to x.

Question

Differentiate y=log(sinx)y=\log(\sin x) with respect to xx, taking log\log to be the natural logarithm.

Step-by-step solution

  1. Identify the composition. The function is a logarithm wrapped around a trigonometric function, so it is lnu\ln u with the inner function u=sinxu=\sin x. That calls for the chain rule rather than any log identity.

  2. Write down the chain rule. For y=lnuy=\ln u, dydx=1ududx.\frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}.

  3. Differentiate the inner function. With u=sinxu=\sin x we get dudx=cosx\dfrac{du}{dx}=\cos x.

  4. Substitute and simplify. dydx=1sinxcosx=cosxsinx=cotx.\frac{dy}{dx}=\frac{1}{\sin x}\cdot\cos x=\frac{\cos x}{\sin x}=\cot x.

  5. Note the domain and sanity-check. The original function needs sinx>0\sin x>0, so the derivative cotx\cot x is stated on those intervals. At x=1x=1 the derivative is cot10.642093\cot 1\approx 0.642093, which matches a numerical difference quotient of ln(sinx)\ln(\sin x) at x=1x=1.

Answer

dydx=cotx\frac{dy}{dx}=\cot x

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