What is the derivative of g(x)=ln(2x^2+4) ?

The derivative of g(x)=ln(2x^2+4) is (2x)/(x^2+2)

What is the first derivative of g(x)=ln(2x^2+4) ?

The first derivative of g(x)=ln(2x^2+4) is (2x)/(x^2+2)

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derivative of

g (x) = ln (2 x^{2} + 4)

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

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{2 x ^{2} + 4} \cdot 4 x

Simplify

\frac{1}{2 x ^{2} + 4} \cdot 4 x : \frac{2 x}{x ^{2} + 2}

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

(6 t - 1) (4 t - 4)^{- 1}

\frac{d}{d x} (x^{4} - 2 a x^{2} - y + 3)

\int_{0}^{\infty} a x e^{- a x} d x

\int 2 x (x^{2} + 10)^{3} d x

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

What is the derivative of g(x)=ln(2x^2+4) ?
The derivative of g(x)=ln(2x^2+4) is (2x)/(x^2+2)
What is the first derivative of g(x)=ln(2x^2+4) ?
The first derivative of g(x)=ln(2x^2+4) is (2x)/(x^2+2)