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

The derivative of (((3x^2))/(2x+4))^3 is (81x^5(x+4))/(2(2x+4)^2(x+2)^2)

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

The first derivative of (((3x^2))/(2x+4))^3 is (81x^5(x+4))/(2(2x+4)^2(x+2)^2)

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derivative of

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

\frac{81 x ^{5} ( x + 4 )}{2 ( 2 x + 4 ) ^{2} ( x + 2 ) ^{2}}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

3 (\frac{3 x ^{2}}{2 x + 4})^{2} \frac{3 x ( x + 4 )}{2 ( x + 2 ) ^{2}} : \frac{81 x ^{5} ( x + 4 )}{2 ( 2 x + 4 ) ^{2} ( x + 2 ) ^{2}}

= \frac{81 x ^{5} ( x + 4 )}{2 ( 2 x + 4 ) ^{2} ( x + 2 ) ^{2}}

x \to 0 lim (x \cdot ln^{2} (x))

x \to 0 lim (\frac{3 e ^{x} - 3}{x})

x \to \infty lim (- 5 x^{6} - 6 x^{4} - x^{3} + 7 x)

\int \frac{( ln ( x ^{9} ) )}{x} d x

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 12 x + 1)

What is the derivative of (((3x^2))/(2x+4))^3 ?
The derivative of (((3x^2))/(2x+4))^3 is (81x^5(x+4))/(2(2x+4)^2(x+2)^2)
What is the first derivative of (((3x^2))/(2x+4))^3 ?
The first derivative of (((3x^2))/(2x+4))^3 is (81x^5(x+4))/(2(2x+4)^2(x+2)^2)