What is the derivative of cos^2(ln(e^{2x+1}(-2x))) ?

The derivative of cos^2(ln(e^{2x+1}(-2x))) is -(sin(2ln(-2e^{1+2x}x))(1+2x))/x

What is the first derivative of cos^2(ln(e^{2x+1}(-2x))) ?

The first derivative of cos^2(ln(e^{2x+1}(-2x))) is -(sin(2ln(-2e^{1+2x}x))(1+2x))/x

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

- \frac{sin ( 2 ln ( - 2 e ^{1 + 2 x} x ) ) ( 1 + 2 x )}{x}

Solution steps

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

Apply the chain rule:

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

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

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

= 2 cos (ln (e^{2 x + 1} (- 2 x))) (- \frac{sin ( ln ( - 2 e ^{2 x + 1} x ) ) ( 2 x + 1 )}{x})

Simplify

2 cos (ln (e^{2 x + 1} (- 2 x))) (- \frac{sin ( ln ( - 2 e ^{2 x + 1} x ) ) ( 2 x + 1 )}{x}) : - \frac{sin ( 2 ln ( - 2 e ^{1 + 2 x} x ) ) ( 1 + 2 x )}{x}

= - \frac{sin ( 2 ln ( - 2 e ^{1 + 2 x} x ) ) ( 1 + 2 x )}{x}

e^{4 t}

x \to 0 lim ((\frac{2}{x})^{x})

x \to 4 + lim (4 x - 4)

f (x) = 2 x^{3} + 3 x^{2} + 5 x, - 4, 3

\frac{\partial}{\partial x} (e^{3 x} sin (3 y))

What is the derivative of cos^2(ln(e^{2x+1}(-2x))) ?
The derivative of cos^2(ln(e^{2x+1}(-2x))) is -(sin(2ln(-2e^{1+2x}x))(1+2x))/x
What is the first derivative of cos^2(ln(e^{2x+1}(-2x))) ?
The first derivative of cos^2(ln(e^{2x+1}(-2x))) is -(sin(2ln(-2e^{1+2x}x))(1+2x))/x