What is the derivative of 8^{4-x^2} ?

The derivative of 8^{4-x^2} is -3ln(2)*2^{-3x^2+13}x

What is the first derivative of 8^{4-x^2} ?

The first derivative of 8^{4-x^2} is -3ln(2)*2^{-3x^2+13}x

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\frac{d}{d x} (8^{4 - x^{2}})

- 3 ln (2) \cdot 2^{- 3 x^{2} + 13} x

Solution steps

\frac{d}{d x} (8^{4 - x^{2}})

Apply exponent rule:

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

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

= \frac{d}{d x} (e^{(4 - x^{2}) l n (8)})

Apply the chain rule:

e^{(4 - x^{2}) l n (8)} \frac{d}{d x} ((4 - x^{2}) ln (8))

= e^{(4 - x^{2}) l n (8)} \frac{d}{d x} ((4 - x^{2}) ln (8))

\frac{d}{d x} ((4 - x^{2}) ln (8)) = - 6 ln (2) x

= e^{(4 - x^{2}) l n (8)} (- 6 ln (2) x)

Simplify

e^{(4 - x^{2}) l n (8)} (- 6 ln (2) x) : - 3 ln (2) \cdot 2^{- 3 x^{2} + 13} x

= - 3 ln (2) \cdot 2^{- 3 x^{2} + 13} x

(x^{2} - 35)^{6}

\frac{d}{d x} (\frac{1}{sin ( x ) + cos ( x )})

y = x^{2} + 2, y = - x, x = 0, x = 1

\int \frac{29}{( x + 1 ) ( x ^{2} + x )} d x

y = \frac{e ^{x}}{9 x ^{2}}

What is the derivative of 8^{4-x^2} ?
The derivative of 8^{4-x^2} is -3ln(2)*2^{-3x^2+13}x
What is the first derivative of 8^{4-x^2} ?
The first derivative of 8^{4-x^2} is -3ln(2)*2^{-3x^2+13}x