What is the derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) ?

The derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) is -4/(x^3)

What is the first derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) ?

The first derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) is -4/(x^3)

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\frac{d}{d x} (\frac{4}{2 x ^{2} - ( + 4 ) ( - 3 ) - ( - 6 ) ( - 2 )})

- \frac{4}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{4}{2 x ^{2} - ( 4 ) ( - 3 ) - ( - 6 ) ( - 2 )})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{1}{2 x ^{2} - 4 ( - 3 ) - ( - 6 ) ( - 2 )})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 4 \frac{d}{d x} ((2 x^{2} - 4 (- 3) - (- 6) (- 2))^{- 1})

Apply the chain rule:

- \frac{1}{( 2 x ^{2} - 4 ( - 3 ) - ( - 6 ) ( - 2 ) ) ^{2}} \frac{d}{d x} (2 x^{2} - 4 (- 3) - (- 6) (- 2))

= - \frac{1}{( 2 x ^{2} - 4 ( - 3 ) - ( - 6 ) ( - 2 ) ) ^{2}} \frac{d}{d x} (2 x^{2} - 4 (- 3) - (- 6) (- 2))

\frac{d}{d x} (2 x^{2} - 4 (- 3) - (- 6) (- 2)) = 4 x

= 4 (- \frac{1}{( 2 x ^{2} - 4 ( - 3 ) - ( - 6 ) ( - 2 ) ) ^{2}} \cdot 4 x)

Simplify

4 (- \frac{1}{( 2 x ^{2} - 4 ( - 3 ) - ( - 6 ) ( - 2 ) ) ^{2}} \cdot 4 x) : - \frac{4}{x ^{3}}

= - \frac{4}{x ^{3}}

\frac{1}{n ( 1 + ln ( x ) ) ^{2}}

\frac{d}{d y} (\frac{y}{2 x + 1})

\frac{1}{n}

(3 x^{2} + 7) (6 x + 3)

(x^{2} - y^{2})^{\frac{1}{2}}

What is the derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) ?
The derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) is -4/(x^3)
What is the first derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) ?
The first derivative of 4/(2x^2-(+4(-3)-(-6)(-2))) is -4/(x^3)