What is the derivative of y=ln((x+3)/(x^2)) ?

The derivative of y=ln((x+3)/(x^2)) is (-x-6)/(x(x+3))

What is the first derivative of y=ln((x+3)/(x^2)) ?

The first derivative of y=ln((x+3)/(x^2)) is (-x-6)/(x(x+3))

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derivative of

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

\frac{- x - 6}{x ( x + 3 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x + 3}{x ^{2}}) = \frac{- x - 6}{x ^{3}}

= \frac{1}{\frac{x + 3}{x ^{2}}} \cdot \frac{- x - 6}{x ^{3}}

Simplify

\frac{1}{\frac{x + 3}{x ^{2}}} \cdot \frac{- x - 6}{x ^{3}} : \frac{- x - 6}{x ( x + 3 )}

= \frac{- x - 6}{x ( x + 3 )}

\frac{d}{d x} (x^{3} 1 - x^{2})

n = 1 \sum \infty \frac{n !}{3 n ! - 1}

y = 3 x + 1, y = 13 + 2 x - x^{2}

f (x) = 2 x + 1, at x = 1

What is the derivative of y=ln((x+3)/(x^2)) ?
The derivative of y=ln((x+3)/(x^2)) is (-x-6)/(x(x+3))
What is the first derivative of y=ln((x+3)/(x^2)) ?
The first derivative of y=ln((x+3)/(x^2)) is (-x-6)/(x(x+3))