What is the derivative of cos^2(ln(x^3+3x)) ?

The derivative of cos^2(ln(x^3+3x)) is -(sin(2ln(x^3+3x))(3+3x^2))/(x^3+3x)

What is the first derivative of cos^2(ln(x^3+3x)) ?

The first derivative of cos^2(ln(x^3+3x)) is -(sin(2ln(x^3+3x))(3+3x^2))/(x^3+3x)

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\frac{d}{d x} (cos^{2} (ln (x^{3} + 3 x)))

- \frac{sin ( 2 ln ( x ^{3} + 3 x ) ) ( 3 + 3 x ^{2} )}{x ^{3} + 3 x}

Solution steps

\frac{d}{d x} (cos^{2} (ln (x^{3} + 3 x)))

Apply the chain rule:

2 cos (ln (x^{3} + 3 x)) \frac{d}{d x} (cos (ln (x^{3} + 3 x)))

= 2 cos (ln (x^{3} + 3 x)) \frac{d}{d x} (cos (ln (x^{3} + 3 x)))

\frac{d}{d x} (cos (ln (x^{3} + 3 x))) = - \frac{sin ( ln ( x ^{3} + 3 x ) ) ( 3 x ^{2} + 3 )}{x ^{3} + 3 x}

= 2 cos (ln (x^{3} + 3 x)) (- \frac{sin ( ln ( x ^{3} + 3 x ) ) ( 3 x ^{2} + 3 )}{x ^{3} + 3 x})

Simplify

2 cos (ln (x^{3} + 3 x)) (- \frac{sin ( ln ( x ^{3} + 3 x ) ) ( 3 x ^{2} + 3 )}{x ^{3} + 3 x}) : - \frac{sin ( 2 ln ( x ^{3} + 3 x ) ) ( 3 + 3 x ^{2} )}{x ^{3} + 3 x}

= - \frac{sin ( 2 ln ( x ^{3} + 3 x ) ) ( 3 + 3 x ^{2} )}{x ^{3} + 3 x}

\int tan^{5} (\frac{x}{2}) d x

\frac{d}{d x} (\frac{( x ^{7} )}{x ^{14} + 1})

\int ln (x^{2} + a^{2}) d x

\frac{\partial}{\partial x} (x + i y)

\int \frac{( x - 1 )}{x ^{3} - x ^{2} - 2 x} d x

What is the derivative of cos^2(ln(x^3+3x)) ?
The derivative of cos^2(ln(x^3+3x)) is -(sin(2ln(x^3+3x))(3+3x^2))/(x^3+3x)
What is the first derivative of cos^2(ln(x^3+3x)) ?
The first derivative of cos^2(ln(x^3+3x)) is -(sin(2ln(x^3+3x))(3+3x^2))/(x^3+3x)