What is the derivative of y=ln(x^4sin^2(x)) ?

The derivative of y=ln(x^4sin^2(x)) is (4sin^2(x)+xsin(2x))/(xsin^2(x))

What is the first derivative of y=ln(x^4sin^2(x)) ?

The first derivative of y=ln(x^4sin^2(x)) is (4sin^2(x)+xsin(2x))/(xsin^2(x))

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derivative of

y = ln (x^{4} sin^{2} (x))

\frac{4 sin ^{2} ( x ) + x sin ( 2 x )}{x sin ^{2} ( x )}

Solution steps

\frac{d}{d x} (ln (x^{4} sin^{2} (x)))

Apply the chain rule:

\frac{1}{x ^{4} sin ^{2} ( x )} \frac{d}{d x} (x^{4} sin^{2} (x))

= \frac{1}{x ^{4} sin ^{2} ( x )} \frac{d}{d x} (x^{4} sin^{2} (x))

\frac{d}{d x} (x^{4} sin^{2} (x)) = 4 x^{3} sin^{2} (x) + sin (2 x) x^{4}

= \frac{1}{x ^{4} sin ^{2} ( x )} (4 x^{3} sin^{2} (x) + sin (2 x) x^{4})

Simplify

\frac{1}{x ^{4} sin ^{2} ( x )} (4 x^{3} sin^{2} (x) + sin (2 x) x^{4}) : \frac{4 sin ^{2} ( x ) + x sin ( 2 x )}{x sin ^{2} ( x )}

= \frac{4 sin ^{2} ( x ) + x sin ( 2 x )}{x sin ^{2} ( x )}

y^{^{'}} = \frac{1}{x}

\int - x e^{3 x} d x

\frac{\partial}{\partial x} (w)

\frac{\partial}{\partial x} (x^{2} e^{x - y})

\frac{d x}{d t} = 2 t e^{2 x}

What is the derivative of y=ln(x^4sin^2(x)) ?
The derivative of y=ln(x^4sin^2(x)) is (4sin^2(x)+xsin(2x))/(xsin^2(x))
What is the first derivative of y=ln(x^4sin^2(x)) ?
The first derivative of y=ln(x^4sin^2(x)) is (4sin^2(x)+xsin(2x))/(xsin^2(x))