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What is the derivative of ln(x^3 • sin x ^2)?

1 Answer

Apr 10, 2018

$\frac{d y}{d x} = \frac{3}{x} + 2 x cos (x^{2}) \frac{1}{sin (x^{2})}$ $dy/dx=3/x+2xcos(x^2)1/(sin(x^2))$

Explanation:

We want to find the derivative of

$y = ln (x^{3} sin (x^{2}))$ $y=ln(x^3sin(x^2))$

Using the properties of logarithms

$y = ln (x^{3}) + ln (sin (x^{2}))$ $y=ln(x^3)+ln(sin(x^2))$

$y = 3 ln (x) + ln (sin (x^{2}))$ $color(white)(y)=3ln(x)+ln(sin(x^2))$

By the chain rule

$\frac{d y}{d x} = \frac{3}{x} + (\frac{d}{d x} (sin (x^{2}))) \frac{1}{sin (x^{2})}$ $dy/dx=3/x+(d/dx(sin(x^2)))1/(sin(x^2))$

$\frac{d y}{d x} = \frac{3}{x} + 2 x cos (x^{2}) \frac{1}{sin (x^{2})}$ $color(white)(dy/dx)=3/x+2xcos(x^2)1/(sin(x^2))$

$\frac{d y}{d x} = \frac{3}{x} + 2 x cot (x^{2})$ $color(white)(dy/dx)=3/x+2xcot(x^2)$

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