What is the derivative of x^{1/(1-x}) ?

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

What is the first derivative of x^{1/(1-x}) ?

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

t y^{^{'}} - 2 y = t^{3} e^{- t}

\int_{0}^{8} 1 + 36 x d x

f (x) = 2 x^{2} + 4 x, at x = - 2

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

\int sin (x) cos (x) + x d x

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