What is the derivative of 1/(sqrt(1+cos^2(2x))) ?

The derivative of 1/(sqrt(1+cos^2(2x))) is (sin(4x))/((1+cos^2(2x))^{3/2)}

What is the first derivative of 1/(sqrt(1+cos^2(2x))) ?

The first derivative of 1/(sqrt(1+cos^2(2x))) is (sin(4x))/((1+cos^2(2x))^{3/2)}

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((1 + cos^{2} (2 x))^{- \frac{1}{2}})

Apply the chain rule:

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

= - \frac{1}{2 ( 1 + cos ^{2} ( 2 x ) ) ^{\frac{3}{2}}} \frac{d}{d x} (1 + cos^{2} (2 x))

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

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

Simplify

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

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

\frac{d}{d x} ((a x^{2} + b x + c) e^{\frac{- x}{8}})

\int \frac{x ^{2} - 2 x + 3}{x ^{3} + 3 x} d x

\frac{\partial}{\partial x} (\frac{1}{y ^{2}})

x d x + y e^{- x} d y = 0, y (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 0

What is the derivative of 1/(sqrt(1+cos^2(2x))) ?
The derivative of 1/(sqrt(1+cos^2(2x))) is (sin(4x))/((1+cos^2(2x))^{3/2)}
What is the first derivative of 1/(sqrt(1+cos^2(2x))) ?
The first derivative of 1/(sqrt(1+cos^2(2x))) is (sin(4x))/((1+cos^2(2x))^{3/2)}