What is the derivative of ln((x^2/((x-4)))) ?

The derivative of ln((x^2/((x-4)))) is (x-8)/(x(x-4))

What is the first derivative of ln((x^2/((x-4)))) ?

The first derivative of ln((x^2/((x-4)))) is (x-8)/(x(x-4))

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\frac{d}{d x} (ln (\frac{x ^{2}}{( x - 4 )}))

\frac{x - 8}{x ( x - 4 )}

Solution steps

\frac{d}{d x} (ln (\frac{x ^{2}}{( x - 4 )}))

Apply the chain rule:

\frac{1}{\frac{x ^{2}}{x - 4}} \frac{d}{d x} (\frac{x ^{2}}{x - 4})

= \frac{1}{\frac{x ^{2}}{x - 4}} \frac{d}{d x} (\frac{x ^{2}}{x - 4})

\frac{d}{d x} (\frac{x ^{2}}{x - 4}) = \frac{x ^{2} - 8 x}{( x - 4 ) ^{2}}

= \frac{1}{\frac{x ^{2}}{x - 4}} \cdot \frac{x ^{2} - 8 x}{( x - 4 ) ^{2}}

Simplify

\frac{1}{\frac{x ^{2}}{x - 4}} \cdot \frac{x ^{2} - 8 x}{( x - 4 ) ^{2}} : \frac{x - 8}{x ( x - 4 )}

= \frac{x - 8}{x ( x - 4 )}

\frac{d}{d x} (x^{3} e^{x^{2}})

f (x) = (3 x - 2)^{4}

\int cos (2 x) - 1 d x

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

\frac{\partial}{\partial x} (x^{2} + z^{2})

What is the derivative of ln((x^2/((x-4)))) ?
The derivative of ln((x^2/((x-4)))) is (x-8)/(x(x-4))
What is the first derivative of ln((x^2/((x-4)))) ?
The first derivative of ln((x^2/((x-4)))) is (x-8)/(x(x-4))