What is the derivative of g(x)=(2x^2+x+9)^{-3} ?

The derivative of g(x)=(2x^2+x+9)^{-3} is -(3(4x+1))/((2x^2+x+9)^4)

What is the first derivative of g(x)=(2x^2+x+9)^{-3} ?

The first derivative of g(x)=(2x^2+x+9)^{-3} is -(3(4x+1))/((2x^2+x+9)^4)

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derivative of

g (x) = (2 x^{2} + x + 9)^{- 3}

- \frac{3 ( 4 x + 1 )}{( 2 x ^{2} + x + 9 ) ^{4}}

Solution steps

\frac{d}{d x} ((2 x^{2} + x + 9)^{- 3})

Apply the chain rule:

- \frac{3}{( 2 x ^{2} + x + 9 ) ^{4}} \frac{d}{d x} (2 x^{2} + x + 9)

= - \frac{3}{( 2 x ^{2} + x + 9 ) ^{4}} \frac{d}{d x} (2 x^{2} + x + 9)

\frac{d}{d x} (2 x^{2} + x + 9) = 4 x + 1

= - \frac{3}{( 2 x ^{2} + x + 9 ) ^{4}} (4 x + 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{3 ( 4 x + 1 )}{( 2 x ^{2} + x + 9 ) ^{4}}

\frac{d}{d y} (e^{3 x} (3 x^{2} y + 2 x y + y^{3}))

\int \frac{cos ( 3 t ) + 3}{sin ( 3 t )} d t

\int \frac{2 ln ( x )}{x ^{3}} d x

\int 2 x + 6 d x

\frac{1}{s ^{4} ( s + 3 )}

What is the derivative of g(x)=(2x^2+x+9)^{-3} ?
The derivative of g(x)=(2x^2+x+9)^{-3} is -(3(4x+1))/((2x^2+x+9)^4)
What is the first derivative of g(x)=(2x^2+x+9)^{-3} ?
The first derivative of g(x)=(2x^2+x+9)^{-3} is -(3(4x+1))/((2x^2+x+9)^4)