What is the derivative of 1/(ln(x+sqrt(1+x^2))) ?

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

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

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial s} (- (s^{2} + t^{2}))

\int_{1}^{e} ln (13 x) d x

\int e^{1 - 8 t} d t

\frac{\partial}{\partial u} (\frac{1}{4} (u - v))

f (x) = \frac{5 ^{4 x^{3}} + e ^{2 x^{3}}}{12}

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