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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Popular >

\frac{d}{d x} ((\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})

\int_{4}^{9} (x) d x

\int \frac{1}{8 - 4 x - 4 x ^{2}} d x

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

\int_{0}^{2} \frac{x}{x ^{2} + 4} d x

x \to - 3 - lim (\frac{1}{x ^{2} - 9})

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))