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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to - 3 lim (f (x))

\int sec^{8} (x) tan (x) d x

\frac{d}{d x} (- x + 2)

f (t) = cos (t) + t^{2} sin (t)

\int - 9 e^{- x} d x

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