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

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

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

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

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

- ln (2) \cdot 2^{- x^{4} + 4} x^{3}

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

\frac{d}{d x} ((2 - x^{4}) ln (2)) = - 4 ln (2) x^{3}

= e^{(2 - x^{4}) l n (2)} (- 4 ln (2) x^{3})

Simplify

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

= - ln (2) \cdot 2^{- x^{4} + 4} x^{3}

\int \frac{e ^{\frac{x ^{2}}{2}}}{x ^{2}} d x

\frac{d}{d x} (3 ln (x^{2} + 1))

x + 2, \frac{1}{1 + x}, x = 2

y^{^{''}} + 7 y^{^{'}} + 10 y = 4 t e^{4 t}

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