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Quotient rule with logarithmic and trigonometric functions

Let h(x)=ln⁡(x)sin⁡(x)h(x) = \dfrac{\ln(x)}{\sin(x)}. Find h′(x)h'(x).

Hint

Use the quotient rule (fg)′=f′g−fg′g2\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2} with f(x)=ln⁡(x)f(x) = \ln(x) and g(x)=sin⁡(x)g(x) = \sin(x). Recall ddx[ln⁡x]=1x\dfrac{d}{dx}[\ln x] = \dfrac{1}{x} and ddx[sin⁡x]=cos⁡x\dfrac{d}{dx}[\sin x] = \cos x.

Step-by-step solution

Identify f(x)=ln⁡(x)f(x) = \ln(x) and g(x)=sin⁡(x)g(x) = \sin(x), so f′(x)=1xf'(x) = \dfrac{1}{x} and g′(x)=cos⁡(x)g'(x) = \cos(x). Apply the quotient rule: h′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2h'(x) = \dfrac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.

Substituting gives h′(x)=1xsin⁡(x)−ln⁡(x)cos⁡(x)sin⁡2(x)h'(x) = \dfrac{\frac{1}{x}\sin(x) - \ln(x)\cos(x)}{\sin^2(x)}. This can be left as is or rewritten with a common denominator inside the numerator as sin⁡(x)−xln⁡(x)cos⁡(x)xsin⁡2(x)\dfrac{\sin(x) - x\ln(x)\cos(x)}{x\sin^2(x)}.

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