What is the derivative of 10^{1-x^2} ?

The derivative of 10^{1-x^2} is -ln(10)*5^{1-x^2}*2^{-x^2+2}x

What is the first derivative of 10^{1-x^2} ?

The first derivative of 10^{1-x^2} is -ln(10)*5^{1-x^2}*2^{-x^2+2}x

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\frac{d}{d x} (1 0^{1 - x^{2}})

- ln (10) \cdot 5^{1 - x^{2}} \cdot 2^{- x^{2} + 2} x

Solution steps

\frac{d}{d x} (1 0^{1 - x^{2}})

Apply exponent rule:

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

1 0^{1 - x^{2}} = e^{(1 - x^{2}) l n (10)}

= \frac{d}{d x} (e^{(1 - x^{2}) l n (10)})

Apply the chain rule:

e^{(1 - x^{2}) l n (10)} \frac{d}{d x} ((1 - x^{2}) ln (10))

= e^{(1 - x^{2}) l n (10)} \frac{d}{d x} ((1 - x^{2}) ln (10))

\frac{d}{d x} ((1 - x^{2}) ln (10)) = - 2 ln (10) x

= e^{(1 - x^{2}) l n (10)} (- 2 ln (10) x)

Simplify

e^{(1 - x^{2}) l n (10)} (- 2 ln (10) x) : - ln (10) \cdot 5^{1 - x^{2}} \cdot 2^{- x^{2} + 2} x

= - ln (10) \cdot 5^{1 - x^{2}} \cdot 2^{- x^{2} + 2} x

\frac{\partial}{\partial x} (arctan (x + 4 y))

x \to - 2 lim (\frac{x ^{2} + 8}{x - 2})

y^{^{''}} = - 3 y - 4 y^{^{'}}

\frac{d y}{d x} = e^{- 2 x + 3 y}

What is the derivative of 10^{1-x^2} ?
The derivative of 10^{1-x^2} is -ln(10)*5^{1-x^2}*2^{-x^2+2}x
What is the first derivative of 10^{1-x^2} ?
The first derivative of 10^{1-x^2} is -ln(10)*5^{1-x^2}*2^{-x^2+2}x