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

The derivative of g(x)=((3x^2-2)/(2x+3))^3 is (3(3x^2-2)^2(6x^2+18x+4))/((2x+3)^4)

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

The first derivative of g(x)=((3x^2-2)/(2x+3))^3 is (3(3x^2-2)^2(6x^2+18x+4))/((2x+3)^4)

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derivative of

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

\frac{3 ( 3 x ^{2} - 2 ) ^{2} ( 6 x ^{2} + 18 x + 4 )}{( 2 x + 3 ) ^{4}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{3 x ^{2} - 2}{2 x + 3}) = \frac{6 x ^{2} + 18 x + 4}{( 2 x + 3 ) ^{2}}

= 3 (\frac{3 x ^{2} - 2}{2 x + 3})^{2} \frac{6 x ^{2} + 18 x + 4}{( 2 x + 3 ) ^{2}}

Simplify

3 (\frac{3 x ^{2} - 2}{2 x + 3})^{2} \frac{6 x ^{2} + 18 x + 4}{( 2 x + 3 ) ^{2}} : \frac{3 ( 3 x ^{2} - 2 ) ^{2} ( 6 x ^{2} + 18 x + 4 )}{( 2 x + 3 ) ^{4}}

= \frac{3 ( 3 x ^{2} - 2 ) ^{2} ( 6 x ^{2} + 18 x + 4 )}{( 2 x + 3 ) ^{4}}

5 x^{2} - 2 x

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{8} e^{- x} d y = 0

\frac{d}{d x} (- a^{2} sin (a x))

\frac{d}{d t} (e^{6 t})

\int x (5 x^{2} + 9)^{8} d x

What is the derivative of g(x)=((3x^2-2)/(2x+3))^3 ?
The derivative of g(x)=((3x^2-2)/(2x+3))^3 is (3(3x^2-2)^2(6x^2+18x+4))/((2x+3)^4)
What is the first derivative of g(x)=((3x^2-2)/(2x+3))^3 ?
The first derivative of g(x)=((3x^2-2)/(2x+3))^3 is (3(3x^2-2)^2(6x^2+18x+4))/((2x+3)^4)