What is the derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) ?

The derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) is (10x^2-40)/((x^2+5x+4)^2)

What is the first derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) ?

The first derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) is (10x^2-40)/((x^2+5x+4)^2)

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derivative of

f (x) = \frac{x ^{2} - 5 x + 4}{x ^{2} + 5 x + 4}

\frac{10 x ^{2} - 40}{( x ^{2} + 5 x + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 5 x + 4}{x ^{2} + 5 x + 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 5 x + 4 ) ( x ^{2} + 5 x + 4 ) - \frac{d}{d x} ( x ^{2} + 5 x + 4 ) ( x ^{2} - 5 x + 4 )}{( x ^{2} + 5 x + 4 ) ^{2}}

\frac{d}{d x} (x^{2} - 5 x + 4) = 2 x - 5

\frac{d}{d x} (x^{2} + 5 x + 4) = 2 x + 5

= \frac{( 2 x - 5 ) ( x ^{2} + 5 x + 4 ) - ( 2 x + 5 ) ( x ^{2} - 5 x + 4 )}{( x ^{2} + 5 x + 4 ) ^{2}}

Expand

(2 x - 5) (x^{2} + 5 x + 4) - (2 x + 5) (x^{2} - 5 x + 4) : 10 x^{2} - 40

= \frac{10 x ^{2} - 40}{( x ^{2} + 5 x + 4 ) ^{2}}

y = x + 5, (4, 3)

\int (- \frac{2000}{x ^{2}}) d x

\int (3 x^{2} + 2 x) \cdot ln (x) d x

x \to - 8 lim (\frac{8 - x}{8 + x})

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

What is the derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) ?
The derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) is (10x^2-40)/((x^2+5x+4)^2)
What is the first derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) ?
The first derivative of f(x)=(x^2-5x+4)/(x^2+5x+4) is (10x^2-40)/((x^2+5x+4)^2)