What is the derivative of log_{x}(x^2+1) ?

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

What is the first derivative of log_{x}(x^2+1) ?

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

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

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

Solution steps

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

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

\int \frac{( x ^{2} + 4 )}{( x - 5 ) ( x ^{2} + 9 )} d x

\int x^{4} e^{(1 - x^{5})} d x

\frac{d y}{d x} + 5 y = 8

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

\int 16 x^{2} d x

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