What is the derivative of 4^{9-x^2} ?

The derivative of 4^{9-x^2} is -ln(2)*4^{-x^2+10}x

What is the first derivative of 4^{9-x^2} ?

The first derivative of 4^{9-x^2} is -ln(2)*4^{-x^2+10}x

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\frac{d}{d x} (4^{9 - x^{2}})

- ln (2) \cdot 4^{- x^{2} + 10} x

Solution steps

\frac{d}{d x} (4^{9 - x^{2}})

Apply exponent rule:

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

4^{9 - x^{2}} = e^{(9 - x^{2}) l n (4)}

= \frac{d}{d x} (e^{(9 - x^{2}) l n (4)})

Apply the chain rule:

e^{(9 - x^{2}) l n (4)} \frac{d}{d x} ((9 - x^{2}) ln (4))

= e^{(9 - x^{2}) l n (4)} \frac{d}{d x} ((9 - x^{2}) ln (4))

\frac{d}{d x} ((9 - x^{2}) ln (4)) = - 4 ln (2) x

= e^{(9 - x^{2}) l n (4)} (- 4 ln (2) x)

Simplify

e^{(9 - x^{2}) l n (4)} (- 4 ln (2) x) : - ln (2) \cdot 4^{- x^{2} + 10} x

= - ln (2) \cdot 4^{- x^{2} + 10} x

\int \frac{1}{1 + y ^{2}} d y

x \to 0 lim (\frac{x ^{2} - 4}{x - 2})

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

\frac{\partial}{\partial u} (ln (4 u^{2} + 5 v^{3}))

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

What is the derivative of 4^{9-x^2} ?
The derivative of 4^{9-x^2} is -ln(2)*4^{-x^2+10}x
What is the first derivative of 4^{9-x^2} ?
The first derivative of 4^{9-x^2} is -ln(2)*4^{-x^2+10}x