What is the derivative of 1/(2x^2-3x-5) ?

The derivative of 1/(2x^2-3x-5) is -(4x-3)/((2x^2-3x-5)^2)

What is the first derivative of 1/(2x^2-3x-5) ?

The first derivative of 1/(2x^2-3x-5) is -(4x-3)/((2x^2-3x-5)^2)

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\frac{d}{d x} (\frac{1}{2 x ^{2} - 3 x - 5})

- \frac{4 x - 3}{( 2 x ^{2} - 3 x - 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{2 x ^{2} - 3 x - 5})

Apply exponent rule:

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

= \frac{d}{d x} ((2 x^{2} - 3 x - 5)^{- 1})

Apply the chain rule:

- \frac{1}{( 2 x ^{2} - 3 x - 5 ) ^{2}} \frac{d}{d x} (2 x^{2} - 3 x - 5)

= - \frac{1}{( 2 x ^{2} - 3 x - 5 ) ^{2}} \frac{d}{d x} (2 x^{2} - 3 x - 5)

\frac{d}{d x} (2 x^{2} - 3 x - 5) = 4 x - 3

= - \frac{1}{( 2 x ^{2} - 3 x - 5 ) ^{2}} (4 x - 3)

Simplify

- \frac{1}{( 2 x ^{2} - 3 x - 5 ) ^{2}} (4 x - 3) : - \frac{4 x - 3}{( 2 x ^{2} - 3 x - 5 ) ^{2}}

= - \frac{4 x - 3}{( 2 x ^{2} - 3 x - 5 ) ^{2}}

\int 3 e^{x} sin (x) d x

4 (t) + ∣ 4 (t) - 1 ∣

\int \frac{1}{( x ^{2} - 6 x + 13 ) ^{2}} d x

\int_{0}^{1} \frac{50}{4 y - 1} d y

{t + e^{2 t}}

What is the derivative of 1/(2x^2-3x-5) ?
The derivative of 1/(2x^2-3x-5) is -(4x-3)/((2x^2-3x-5)^2)
What is the first derivative of 1/(2x^2-3x-5) ?
The first derivative of 1/(2x^2-3x-5) is -(4x-3)/((2x^2-3x-5)^2)