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

The derivative of f(x)=(x^3)/((x-2)^2) is (x^2(x-6))/((x-2)^3)

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

The first derivative of f(x)=(x^3)/((x-2)^2) is (x^2(x-6))/((x-2)^3)

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derivative of

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

\frac{x ^{2} ( x - 6 )}{( x - 2 ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{3 x ^{2} ( x - 2 ) ^{2} - 2 ( x - 2 ) x ^{3}}{( ( x - 2 ) ^{2} ) ^{2}} : \frac{x ^{2} ( x - 6 )}{( x - 2 ) ^{3}}

= \frac{x ^{2} ( x - 6 )}{( x - 2 ) ^{3}}

\int sec^{3} (θ) d θ

y = sin (5 x) + sin^{2} (5 x), (0, 0)

\int_{0}^{1} \frac{3 x}{e ^{2 x}} d x

sin (t) H (t - π)

What is the derivative of f(x)=(x^3)/((x-2)^2) ?
The derivative of f(x)=(x^3)/((x-2)^2) is (x^2(x-6))/((x-2)^3)
What is the first derivative of f(x)=(x^3)/((x-2)^2) ?
The first derivative of f(x)=(x^3)/((x-2)^2) is (x^2(x-6))/((x-2)^3)