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

The derivative of f(x)=(x^2-4)/(x-3) is (x^2-6x+4)/((x-3)^2)

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

The first derivative of f(x)=(x^2-4)/(x-3) is (x^2-6x+4)/((x-3)^2)

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derivative of

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

\frac{x ^{2} - 6 x + 4}{( x - 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x - 3) = 1

= \frac{2 x ( x - 3 ) - 1 \cdot ( x ^{2} - 4 )}{( x - 3 ) ^{2}}

Simplify

\frac{2 x ( x - 3 ) - 1 \cdot ( x ^{2} - 4 )}{( x - 3 ) ^{2}} : \frac{x ^{2} - 6 x + 4}{( x - 3 ) ^{2}}

= \frac{x ^{2} - 6 x + 4}{( x - 3 ) ^{2}}

x \to - \infty lim (a^{x})

\int y^{2} ln (y) d y

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

x \to 0 lim (x + x)

\int 9 tan^{2} (x) sec^{3} (x) d x

What is the derivative of f(x)=(x^2-4)/(x-3) ?
The derivative of f(x)=(x^2-4)/(x-3) is (x^2-6x+4)/((x-3)^2)
What is the first derivative of f(x)=(x^2-4)/(x-3) ?
The first derivative of f(x)=(x^2-4)/(x-3) is (x^2-6x+4)/((x-3)^2)