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

The derivative of (x/(x-1))^x is (x/(x-1))^x(ln(x/(x-1))-1/(x-1))

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

The first derivative of (x/(x-1))^x is (x/(x-1))^x(ln(x/(x-1))-1/(x-1))

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

y^{^{''}} + 4 y^{^{'}} - 45 y = 0

f (x) = 3.6 x + 500

\int \frac{1}{2 - e ^{3 x}} d x

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

n = 1 \sum \infty \frac{3}{n ^{2} + 5 n + 4}

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