What is the derivative of log_{10}(x/(sqrt(x^2+1))) ?

The derivative of log_{10}(x/(sqrt(x^2+1))) is 1/(ln(10)x(x^2+1))

What is the first derivative of log_{10}(x/(sqrt(x^2+1))) ?

The first derivative of log_{10}(x/(sqrt(x^2+1))) is 1/(ln(10)x(x^2+1))

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\frac{d}{d x} (lo g_{10} (\frac{x}{x ^{2} + 1}))

\frac{1}{ln ( 10 ) x ( x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{10} (\frac{x}{x ^{2} + 1}))

Apply log rule:

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

= \frac{d}{d x} \frac{ln ( \frac{x}{x ^{2} + 1} )}{ln ( 10 )}

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (\frac{x}{x ^{2} + 1}))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 10 )} \cdot \frac{1}{\frac{x}{x ^{2} + 1}} \cdot \frac{1}{( x ^{2} + 1 ) x ^{2} + 1}

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{\frac{x}{x ^{2} + 1}} \cdot \frac{1}{( x ^{2} + 1 ) x ^{2} + 1} : \frac{1}{ln ( 10 ) x ( x ^{2} + 1 )}

= \frac{1}{ln ( 10 ) x ( x ^{2} + 1 )}

\frac{\partial}{\partial x} (15 - 2 x^{2} - y^{2})

(3 + x^{4})^{\frac{4}{5}}

x \to 0 + lim (\frac{10}{x ^{2}})

f (x) = 17 e^{x} - e x^{e}

f (x) = \frac{x}{x + 2}

What is the derivative of log_{10}(x/(sqrt(x^2+1))) ?
The derivative of log_{10}(x/(sqrt(x^2+1))) is 1/(ln(10)x(x^2+1))
What is the first derivative of log_{10}(x/(sqrt(x^2+1))) ?
The first derivative of log_{10}(x/(sqrt(x^2+1))) is 1/(ln(10)x(x^2+1))