What is the derivative of x^{-2cos(x}) ?

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

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

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

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\frac{d}{d x} (x^{- 2 c o s (x)})

x^{- 2 c o s (x)} (2 sin (x) ln (x) - \frac{2 cos ( x )}{x})

Solution steps

\frac{d}{d x} (x^{- 2 c o s (x)})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{- 2 c o s (x)} = e^{(- 2 c o s (x)) l n (x)}

= \frac{d}{d x} (e^{(- 2 c o s (x)) l n (x)})

Apply the chain rule:

e^{(- 2 c o s (x)) l n (x)} \frac{d}{d x} ((- 2 cos (x)) ln (x))

= e^{(- 2 c o s (x)) l n (x)} \frac{d}{d x} ((- 2 cos (x)) ln (x))

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

= e^{(- 2 c o s (x)) l n (x)} (2 sin (x) ln (x) - \frac{2 cos ( x )}{x})

Simplify

e^{(- 2 c o s (x)) l n (x)} (2 sin (x) ln (x) - \frac{2 cos ( x )}{x}) : x^{- 2 c o s (x)} (2 sin (x) ln (x) - \frac{2 cos ( x )}{x})

= x^{- 2 c o s (x)} (2 sin (x) ln (x) - \frac{2 cos ( x )}{x})

x \to 0 lim (\frac{( e ^{x} - e ^{- x} - 2 x )}{1 - sin ( x )})

x \to \infty lim (x \cdot π)

x \to \infty lim (\frac{1}{6})

y^{^{'}} = 1 + x^{2}

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

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