What is the derivative of 4^{-x^8} ?

The derivative of 4^{-x^8} is -ln(2)*2^{-2x^8+4}x^7

What is the first derivative of 4^{-x^8} ?

The first derivative of 4^{-x^8} is -ln(2)*2^{-2x^8+4}x^7

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derivative of

4^{- x^{8}}

- ln (2) \cdot 2^{- 2 x^{8} + 4} x^{7}

Solution steps

\frac{d}{d x} (4^{- x^{8}})

Apply exponent rule:

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

4^{- x^{8}} = e^{(- x^{8}) l n (4)}

= \frac{d}{d x} (e^{(- x^{8}) l n (4)})

Apply the chain rule:

e^{(- x^{8}) l n (4)} \frac{d}{d x} ((- x^{8}) ln (4))

= e^{(- x^{8}) l n (4)} \frac{d}{d x} ((- x^{8}) ln (4))

\frac{d}{d x} ((- x^{8}) ln (4)) = - 16 ln (2) x^{7}

= e^{(- x^{8}) l n (4)} (- 16 ln (2) x^{7})

Simplify

e^{(- x^{8}) l n (4)} (- 16 ln (2) x^{7}) : - ln (2) \cdot 2^{- 2 x^{8} + 4} x^{7}

= - ln (2) \cdot 2^{- 2 x^{8} + 4} x^{7}

\int 2 sin^{2} (x) d x

x^{3} + y^{2} x^{4} + 4 y^{^{'}} = 0

\int_{0}^{x} (e^{- t^{2}}) d t

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

\frac{d}{d x} (\frac{f ( x )}{g ( x ) - 2})

What is the derivative of 4^{-x^8} ?
The derivative of 4^{-x^8} is -ln(2)*2^{-2x^8+4}x^7
What is the first derivative of 4^{-x^8} ?
The first derivative of 4^{-x^8} is -ln(2)*2^{-2x^8+4}x^7