What is the derivative of \sqrt[x]{(x+1)^2} ?

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

What is the first derivative of \sqrt[x]{(x+1)^2} ?

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

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derivative of

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

Solution steps

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

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

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

\frac{∣ x ∣}{5 - x ^{2}}

\frac{d}{d x} (6 - e^{- x})

\frac{d y}{d x} = 7 y e^{x + 3}

\frac{d}{d x} ((2^{x})^{8})

\frac{d}{d x} (x^{2} + e^{x})

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