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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (e^{a x^{4}})

\int_{0}^{1} cos (2 x) d x

\frac{\partial}{\partial y} (\frac{5}{2} x^{2} + 4 x y)

\int \frac{t}{1 - t ^{8}} d t

\frac{\partial}{\partial x} ((x - y)^{10})

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