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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} ((3 - 2 x)^{4}) = - 8 (3 - 2 x)^{3}

= 2 \cdot \frac{2 x ( 3 - 2 x ) ^{4} - ( - 8 ( 3 - 2 x ) ^{3} ) x ^{2}}{( ( 3 - 2 x ) ^{4} ) ^{2}}

Simplify

2 \cdot \frac{2 x ( 3 - 2 x ) ^{4} - ( - 8 ( 3 - 2 x ) ^{3} ) x ^{2}}{( ( 3 - 2 x ) ^{4} ) ^{2}} : \frac{4 x ( 2 x + 3 )}{( 3 - 2 x ) ^{5}}

= \frac{4 x ( 2 x + 3 )}{( 3 - 2 x ) ^{5}}

\int \frac{sec ( θ )}{tan ^{2} ( θ )} d θ

\frac{d}{d θ} (cot^{2} (cos (θ)))

\frac{x + 1 - 3}{x - 8}

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

\int_{0}^{\frac{π}{2}} 2 cos^{5} (x) d x

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