What is the derivative of (3x-1/(sqrt(x^2+2))) ?

The derivative of (3x-1/(sqrt(x^2+2))) is (x+6)/((x^2+2)sqrt(x^2+2))

What is the first derivative of (3x-1/(sqrt(x^2+2))) ?

The first derivative of (3x-1/(sqrt(x^2+2))) is (x+6)/((x^2+2)sqrt(x^2+2))

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

\frac{x + 6}{( x ^{2} + 2 ) x ^{2} + 2}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (3 x - 1) = 3

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

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

Simplify

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

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

What is the derivative of (3x-1/(sqrt(x^2+2))) ?
The derivative of (3x-1/(sqrt(x^2+2))) is (x+6)/((x^2+2)sqrt(x^2+2))
What is the first derivative of (3x-1/(sqrt(x^2+2))) ?
The first derivative of (3x-1/(sqrt(x^2+2))) is (x+6)/((x^2+2)sqrt(x^2+2))