What is the derivative of y=x^{-5cos(x)} ?

The derivative of y=x^{-5cos(x)} is x^{-5cos(x)}(5sin(x)ln(x)-(5cos(x))/x)

What is the first derivative of y=x^{-5cos(x)} ?

The first derivative of y=x^{-5cos(x)} is x^{-5cos(x)}(5sin(x)ln(x)-(5cos(x))/x)

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derivative of

y = x^{- 5 c o s (x)}

x^{- 5 c o s (x)} (5 sin (x) ln (x) - \frac{5 cos ( x )}{x})

Solution steps

\frac{d}{d x} (x^{- 5 c o s (x)})

Apply exponent rule:

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

x^{- 5 c o s (x)} = e^{(- 5 c o s (x)) l n (x)}

= \frac{d}{d x} (e^{(- 5 c o s (x)) l n (x)})

Apply the chain rule:

e^{(- 5 c o s (x)) l n (x)} \frac{d}{d x} ((- 5 cos (x)) ln (x))

= e^{(- 5 c o s (x)) l n (x)} \frac{d}{d x} ((- 5 cos (x)) ln (x))

\frac{d}{d x} ((- 5 cos (x)) ln (x)) = 5 sin (x) ln (x) - \frac{5 cos ( x )}{x}

= e^{(- 5 c o s (x)) l n (x)} (5 sin (x) ln (x) - \frac{5 cos ( x )}{x})

Simplify

e^{(- 5 c o s (x)) l n (x)} (5 sin (x) ln (x) - \frac{5 cos ( x )}{x}) : x^{- 5 c o s (x)} (5 sin (x) ln (x) - \frac{5 cos ( x )}{x})

= x^{- 5 c o s (x)} (5 sin (x) ln (x) - \frac{5 cos ( x )}{x})

x ln (x) - x

\int \frac{x + 4}{x ^{2} + 6 x - 16} d x

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

y = cos (x), at x = - \frac{π}{2}

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

What is the derivative of y=x^{-5cos(x)} ?
The derivative of y=x^{-5cos(x)} is x^{-5cos(x)}(5sin(x)ln(x)-(5cos(x))/x)
What is the first derivative of y=x^{-5cos(x)} ?
The first derivative of y=x^{-5cos(x)} is x^{-5cos(x)}(5sin(x)ln(x)-(5cos(x))/x)