What is the derivative of f(x)=(2x)/(sqrt(x^2+2)) ?

The derivative of f(x)=(2x)/(sqrt(x^2+2)) is 4/((x^2+2)sqrt(x^2+2))

What is the first derivative of f(x)=(2x)/(sqrt(x^2+2)) ?

The first derivative of f(x)=(2x)/(sqrt(x^2+2)) is 4/((x^2+2)sqrt(x^2+2))

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derivative of

f (x) = \frac{2 x}{x ^{2} + 2}

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

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