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

The derivative of (x^2-2x+5/(x^2+2x-3)) is (4x^2-16x-4)/((x^2+2x-3)^2)

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

The first derivative of (x^2-2x+5/(x^2+2x-3)) is (4x^2-16x-4)/((x^2+2x-3)^2)

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

\frac{4 x ^{2} - 16 x - 4}{( x ^{2} + 2 x - 3 ) ^{2}}

Solution steps

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

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

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

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

Expand

(2 x - 2) (x^{2} + 2 x - 3) - (2 x + 2) (x^{2} - 2 x + 5) : 4 x^{2} - 16 x - 4

= \frac{4 x ^{2} - 16 x - 4}{( x ^{2} + 2 x - 3 ) ^{2}}

\int \frac{4 x ^{5} - 4}{x ^{2} - 4} d x

n = 1 \sum \infty \frac{n}{1 + n ^{2}}

\frac{\partial}{\partial y} (ln (e^{x} + e^{y}))

y^{^{''}} + 8 y^{^{'}} + 41 y = 0

arccos (1 - x^{2})

What is the derivative of (x^2-2x+5/(x^2+2x-3)) ?
The derivative of (x^2-2x+5/(x^2+2x-3)) is (4x^2-16x-4)/((x^2+2x-3)^2)
What is the first derivative of (x^2-2x+5/(x^2+2x-3)) ?
The first derivative of (x^2-2x+5/(x^2+2x-3)) is (4x^2-16x-4)/((x^2+2x-3)^2)