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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

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

\int_{0}^{5} \frac{1}{x ^{\frac{14}{13}}} d x

f (x) = 3 x^{5} e^{2 x}

\frac{d}{d x} (9 x + 8)

4 x^{6} (x^{4} - 3 x)

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