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

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

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

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

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derivative of

y = (\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 2 + lim (1)

\int x + x^{2} + 2 d x

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

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

\int \frac{1}{x ^{4} ( x ^{2} + 3 ) ^{\frac{1}{2}}} d x

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