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

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

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

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

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derivative of

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(x + 10)^{\frac{1}{4}}

5 q^{2}

\frac{6}{4 x}

\frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t - 1

6^{x + 2}

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