What is the derivative of f(x)=(x^3+2x+7)/(x^2-5) ?

The derivative of f(x)=(x^3+2x+7)/(x^2-5) is (x^4-17x^2-14x-10)/((x^2-5)^2)

What is the first derivative of f(x)=(x^3+2x+7)/(x^2-5) ?

The first derivative of f(x)=(x^3+2x+7)/(x^2-5) is (x^4-17x^2-14x-10)/((x^2-5)^2)

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derivative of

f (x) = \frac{x ^{3} + 2 x + 7}{x ^{2} - 5}

\frac{x ^{4} - 17 x ^{2} - 14 x - 10}{( x ^{2} - 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 2 x + 7}{x ^{2} - 5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} + 2 x + 7 ) ( x ^{2} - 5 ) - \frac{d}{d x} ( x ^{2} - 5 ) ( x ^{3} + 2 x + 7 )}{( x ^{2} - 5 ) ^{2}}

\frac{d}{d x} (x^{3} + 2 x + 7) = 3 x^{2} + 2

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

= \frac{( 3 x ^{2} + 2 ) ( x ^{2} - 5 ) - 2 x ( x ^{3} + 2 x + 7 )}{( x ^{2} - 5 ) ^{2}}

Expand

(3 x^{2} + 2) (x^{2} - 5) - 2 x (x^{3} + 2 x + 7) : x^{4} - 17 x^{2} - 14 x - 10

= \frac{x ^{4} - 17 x ^{2} - 14 x - 10}{( x ^{2} - 5 ) ^{2}}

\int_{1}^{1.5} \frac{2}{5} (x + 1) d x

f (t) = 6 t^{3} - 54 t^{2} + 70

(10 - 6 y + e^{- 3 x}) d x - 2 d y = 0

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} + x + 1})

What is the derivative of f(x)=(x^3+2x+7)/(x^2-5) ?
The derivative of f(x)=(x^3+2x+7)/(x^2-5) is (x^4-17x^2-14x-10)/((x^2-5)^2)
What is the first derivative of f(x)=(x^3+2x+7)/(x^2-5) ?
The first derivative of f(x)=(x^3+2x+7)/(x^2-5) is (x^4-17x^2-14x-10)/((x^2-5)^2)