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

The derivative of y=(x^2)/(x-4) is (x^2-8x)/((x-4)^2)

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

The first derivative of y=(x^2)/(x-4) is (x^2-8x)/((x-4)^2)

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derivative of

y = \frac{x ^{2}}{x - 4}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x - 4) = 1

= \frac{2 x ( x - 4 ) - 1 \cdot x ^{2}}{( x - 4 ) ^{2}}

Simplify

\frac{2 x ( x - 4 ) - 1 \cdot x ^{2}}{( x - 4 ) ^{2}} : \frac{x ^{2} - 8 x}{( x - 4 ) ^{2}}

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

What is the derivative of y=(x^2)/(x-4) ?
The derivative of y=(x^2)/(x-4) is (x^2-8x)/((x-4)^2)
What is the first derivative of y=(x^2)/(x-4) ?
The first derivative of y=(x^2)/(x-4) is (x^2-8x)/((x-4)^2)