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

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

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

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

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derivative of

y = \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}}

f (x) = 9 - 2 x^{2}, (2, 1)

x \to - 5 lim (\frac{x ^{2} + 3 x - 10}{2 x ^{2} + 15 x + 25})

\int 8 sin^{3} (x) cos^{2} (x) d x

f (x) = (1 + x + x^{2})^{99}

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