What is the derivative of y=ln(9+x^2) ?

The derivative of y=ln(9+x^2) is (2x)/(9+x^2)

What is the first derivative of y=ln(9+x^2) ?

The first derivative of y=ln(9+x^2) is (2x)/(9+x^2)

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derivative of

y = ln (9 + x^{2})

\frac{2 x}{9 + x ^{2}}

Solution steps

\frac{d}{d x} (ln (9 + x^{2}))

Apply the chain rule:

\frac{1}{9 + x ^{2}} \frac{d}{d x} (9 + x^{2})

= \frac{1}{9 + x ^{2}} \frac{d}{d x} (9 + x^{2})

\frac{d}{d x} (9 + x^{2}) = 2 x

= \frac{1}{9 + x ^{2}} \cdot 2 x

Simplify

\frac{1}{9 + x ^{2}} \cdot 2 x : \frac{2 x}{9 + x ^{2}}

= \frac{2 x}{9 + x ^{2}}

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

\int x^{- 13} d x

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\int (\frac{1}{x ( x - 4 )}) d x

\int \frac{2 x ^{2}}{1 - 2 x ^{3}} d x

What is the derivative of y=ln(9+x^2) ?
The derivative of y=ln(9+x^2) is (2x)/(9+x^2)
What is the first derivative of y=ln(9+x^2) ?
The first derivative of y=ln(9+x^2) is (2x)/(9+x^2)