What is the derivative of f(x)=(2-3x-x^2)/(x^2-5) ?

The derivative of f(x)=(2-3x-x^2)/(x^2-5) is (3x^2+6x+15)/((x^2-5)^2)

What is the first derivative of f(x)=(2-3x-x^2)/(x^2-5) ?

The first derivative of f(x)=(2-3x-x^2)/(x^2-5) is (3x^2+6x+15)/((x^2-5)^2)

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derivative of

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

\frac{3 x ^{2} + 6 x + 15}{( x ^{2} - 5 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{2} - 5) = 2 x

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

Expand

(- 2 x - 3) (x^{2} - 5) - 2 x (2 - 3 x - x^{2}) : 3 x^{2} + 6 x + 15

= \frac{3 x ^{2} + 6 x + 15}{( x ^{2} - 5 ) ^{2}}

\int x^{e} + \frac{2}{x} + sec^{2} (2 x) d x

x \to 0 lim (\frac{cos ( 7 x ) tan ( 7 x )}{x})

f (x) = (1 + x) cos (x), at x = 0

y = \frac{1}{3} x^{2} - 2 x + 4, (3, 1)

\frac{d}{d x} (cos (tan (4 x)))

What is the derivative of f(x)=(2-3x-x^2)/(x^2-5) ?
The derivative of f(x)=(2-3x-x^2)/(x^2-5) is (3x^2+6x+15)/((x^2-5)^2)
What is the first derivative of f(x)=(2-3x-x^2)/(x^2-5) ?
The first derivative of f(x)=(2-3x-x^2)/(x^2-5) is (3x^2+6x+15)/((x^2-5)^2)