What is the derivative of log_{3}((x^2/(x+1))) ?

The derivative of log_{3}((x^2/(x+1))) is (x+2)/(ln(3)x(x+1))

What is the first derivative of log_{3}((x^2/(x+1))) ?

The first derivative of log_{3}((x^2/(x+1))) is (x+2)/(ln(3)x(x+1))

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\frac{d}{d x} (lo g_{3} (\frac{x ^{2}}{x + 1}))

\frac{x + 2}{ln ( 3 ) x ( x + 1 )}

Solution steps

\frac{d}{d x} (lo g_{3} (\frac{x ^{2}}{x + 1}))

Apply log rule:

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

= \frac{d}{d x} \frac{ln ( \frac{x ^{2}}{x + 1} )}{ln ( 3 )}

Take the constant out:

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

= \frac{1}{ln ( 3 )} \frac{d}{d x} (ln (\frac{x ^{2}}{x + 1}))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 3 )} \cdot \frac{1}{\frac{x ^{2}}{x + 1}} \cdot \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}}

Simplify

\frac{1}{ln ( 3 )} \cdot \frac{1}{\frac{x ^{2}}{x + 1}} \cdot \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}} : \frac{x + 2}{ln ( 3 ) x ( x + 1 )}

= \frac{x + 2}{ln ( 3 ) x ( x + 1 )}

\frac{d y}{d x} = - 4 (y - 3)

\frac{\partial}{\partial x} (2 x + y^{6})

\int sin^{2} (\frac{x}{2}) d x

f (x) = ln (x + 3 + x^{2})

\int - \frac{x}{y ^{2}} d y

What is the derivative of log_{3}((x^2/(x+1))) ?
The derivative of log_{3}((x^2/(x+1))) is (x+2)/(ln(3)x(x+1))
What is the first derivative of log_{3}((x^2/(x+1))) ?
The first derivative of log_{3}((x^2/(x+1))) is (x+2)/(ln(3)x(x+1))