What is the derivative of g(x)=(1-x^2)^{-x} ?

The derivative of g(x)=(1-x^2)^{-x} is (-ln(1-x^2)+(2x^2)/(1-x^2))(1-x^2)^{-x}

What is the first derivative of g(x)=(1-x^2)^{-x} ?

The first derivative of g(x)=(1-x^2)^{-x} is (-ln(1-x^2)+(2x^2)/(1-x^2))(1-x^2)^{-x}

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derivative of

g (x) = (1 - x^{2})^{- x}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

What is the derivative of g(x)=(1-x^2)^{-x} ?
The derivative of g(x)=(1-x^2)^{-x} is (-ln(1-x^2)+(2x^2)/(1-x^2))(1-x^2)^{-x}
What is the first derivative of g(x)=(1-x^2)^{-x} ?
The first derivative of g(x)=(1-x^2)^{-x} is (-ln(1-x^2)+(2x^2)/(1-x^2))(1-x^2)^{-x}