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 d y - y d x = (x y)^{\frac{1}{2}} d x

f (x) = 7 cos (x), at (\frac{2 π}{3})

x \to 1 lim (\frac{2 - x}{x - 1})

\frac{d}{d x} (x \cdot 2)

(10.1) (5.2)

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)