What is the derivative of log_{7}(x^2+1) ?

The derivative of log_{7}(x^2+1) is (2x)/(ln(7)(x^2+1))

What is the first derivative of log_{7}(x^2+1) ?

The first derivative of log_{7}(x^2+1) is (2x)/(ln(7)(x^2+1))

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\frac{d}{d x} (lo g_{7} (x^{2} + 1))

\frac{2 x}{ln ( 7 ) ( x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{7} (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 ( 7 )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 7 )} \frac{d}{d x} (ln (x^{2} + 1))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 7 )} \cdot \frac{1}{x ^{2} + 1} \cdot 2 x

Simplify

\frac{1}{ln ( 7 )} \cdot \frac{1}{x ^{2} + 1} \cdot 2 x : \frac{2 x}{ln ( 7 ) ( x ^{2} + 1 )}

= \frac{2 x}{ln ( 7 ) ( x ^{2} + 1 )}

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

n = 0 \sum \infty \frac{1}{n ^{1.01}}

x \to 5 lim (x^{2} - 3 x)

x \to 2.99 lim (1 + x)

\int \frac{1}{1 + 1 + x} d x

What is the derivative of log_{7}(x^2+1) ?
The derivative of log_{7}(x^2+1) is (2x)/(ln(7)(x^2+1))
What is the first derivative of log_{7}(x^2+1) ?
The first derivative of log_{7}(x^2+1) is (2x)/(ln(7)(x^2+1))