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

The derivative of g(x)=(3x+1)/((2x^2-5)^2) is (-18x^2-8x-15)/((2x^2-5)^3)

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

The first derivative of g(x)=(3x+1)/((2x^2-5)^2) is (-18x^2-8x-15)/((2x^2-5)^3)

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derivative of

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

\frac{- 18 x ^{2} - 8 x - 15}{( 2 x ^{2} - 5 ) ^{3}}

Solution steps

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

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

\frac{d}{d x} ((2 x^{2} - 5)^{2}) = 8 x (2 x^{2} - 5)

= \frac{3 ( 2 x ^{2} - 5 ) ^{2} - 8 x ( 2 x ^{2} - 5 ) ( 3 x + 1 )}{( ( 2 x ^{2} - 5 ) ^{2} ) ^{2}}

Simplify

\frac{3 ( 2 x ^{2} - 5 ) ^{2} - 8 x ( 2 x ^{2} - 5 ) ( 3 x + 1 )}{( ( 2 x ^{2} - 5 ) ^{2} ) ^{2}} : \frac{- 18 x ^{2} - 8 x - 15}{( 2 x ^{2} - 5 ) ^{3}}

= \frac{- 18 x ^{2} - 8 x - 15}{( 2 x ^{2} - 5 ) ^{3}}

(1 + e^{t})^{2}

\frac{\partial}{\partial x} (8 + x ln (x y - 5))

\int \frac{x ^{2}}{u} d u

f (x) = x^{2} arctan (4 x)

\frac{\partial}{\partial y} (e^{x} + y^{2})

What is the derivative of g(x)=(3x+1)/((2x^2-5)^2) ?
The derivative of g(x)=(3x+1)/((2x^2-5)^2) is (-18x^2-8x-15)/((2x^2-5)^3)
What is the first derivative of g(x)=(3x+1)/((2x^2-5)^2) ?
The first derivative of g(x)=(3x+1)/((2x^2-5)^2) is (-18x^2-8x-15)/((2x^2-5)^3)