What is the derivative of f(x)=arctan(sin(4x)) ?

The derivative of f(x)=arctan(sin(4x)) is (4cos(4x))/(sin^2(4x)+1)

What is the first derivative of f(x)=arctan(sin(4x)) ?

The first derivative of f(x)=arctan(sin(4x)) is (4cos(4x))/(sin^2(4x)+1)

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derivative of

f (x) = arctan (sin (4 x))

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

Solution steps

\frac{d}{d x} (arctan (sin (4 x)))

Apply the chain rule:

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

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

\frac{d}{d x} (sin (4 x)) = cos (4 x) \cdot 4

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

Simplify

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

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

\int \frac{1}{x ^{2} + 2 x + 82} d x

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

\int_{1}^{e^{2}} x^{5} ln (x) d x

\int x y e^{x y^{2}} d y

\frac{d}{d x} (\frac{6 x}{x ^{2} - 4})

What is the derivative of f(x)=arctan(sin(4x)) ?
The derivative of f(x)=arctan(sin(4x)) is (4cos(4x))/(sin^2(4x)+1)
What is the first derivative of f(x)=arctan(sin(4x)) ?
The first derivative of f(x)=arctan(sin(4x)) is (4cos(4x))/(sin^2(4x)+1)