What is the derivative of f(x)=(2x^2)/(x^2-2x-8) ?

The derivative of f(x)=(2x^2)/(x^2-2x-8) is (2(-2x^2-16x))/((x^2-2x-8)^2)

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

The first derivative of f(x)=(2x^2)/(x^2-2x-8) is (2(-2x^2-16x))/((x^2-2x-8)^2)

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derivative of

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

\frac{2 ( - 2 x ^{2} - 16 x )}{( x ^{2} - 2 x - 8 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

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

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

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

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

Simplify

2 \cdot \frac{2 x ( x ^{2} - 2 x - 8 ) - ( 2 x - 2 ) x ^{2}}{( x ^{2} - 2 x - 8 ) ^{2}} : \frac{2 ( - 2 x ^{2} - 16 x )}{( x ^{2} - 2 x - 8 ) ^{2}}

= \frac{2 ( - 2 x ^{2} - 16 x )}{( x ^{2} - 2 x - 8 ) ^{2}}

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

(\frac{t ^{3}}{t ^{8} + 1})^{2}

y^{^{''}} + 4 y = cos (t)

x \to 6 lim (8 x^{2} - 3 x + 1)

What is the derivative of f(x)=(2x^2)/(x^2-2x-8) ?
The derivative of f(x)=(2x^2)/(x^2-2x-8) is (2(-2x^2-16x))/((x^2-2x-8)^2)
What is the first derivative of f(x)=(2x^2)/(x^2-2x-8) ?
The first derivative of f(x)=(2x^2)/(x^2-2x-8) is (2(-2x^2-16x))/((x^2-2x-8)^2)