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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

Simplify

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

\frac{\partial}{\partial u} (2 sin (u) sin (v))

\int x^{2} sin (n x) d x

n = 0 \sum \infty (- 2)^{n}

\frac{\partial}{\partial x} (cos (4 y) e^{x})

x \to 2 lim (\frac{5 x - 25}{x - 2})

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