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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\frac{d}{d x} ((\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}}

\int e^{l n ∣ t ∣} d t

\int \frac{1}{S} d S

f (x) = \frac{( x - 4 )}{x - 2}

\frac{d}{d x} (cos (x) + x)

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

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)