What is the derivative of f(x)=(x^2-1)/((x^2+x+1)) ?

The derivative of f(x)=(x^2-1)/((x^2+x+1)) is (x^2+4x+1)/((x^2+x+1)^2)

What is the first derivative of f(x)=(x^2-1)/((x^2+x+1)) ?

The first derivative of f(x)=(x^2-1)/((x^2+x+1)) is (x^2+4x+1)/((x^2+x+1)^2)

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derivative of

f (x) = \frac{x ^{2} - 1}{( x ^{2} + x + 1 )}

\frac{x ^{2} + 4 x + 1}{( x ^{2} + x + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{2} - 1) = 2 x

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

= \frac{2 x ( x ^{2} + x + 1 ) - ( 2 x + 1 ) ( x ^{2} - 1 )}{( x ^{2} + x + 1 ) ^{2}}

Expand

2 x (x^{2} + x + 1) - (2 x + 1) (x^{2} - 1) : x^{2} + 4 x + 1

= \frac{x ^{2} + 4 x + 1}{( x ^{2} + x + 1 ) ^{2}}

\int ln (5)

y^{^{''}} + 16 y = 3 sin (5 t), y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{1 - 6 x}{3 + x})

x \to 0 - lim (\frac{1}{2 x})

\int (1 - x)^{3} d x

What is the derivative of f(x)=(x^2-1)/((x^2+x+1)) ?
The derivative of f(x)=(x^2-1)/((x^2+x+1)) is (x^2+4x+1)/((x^2+x+1)^2)
What is the first derivative of f(x)=(x^2-1)/((x^2+x+1)) ?
The first derivative of f(x)=(x^2-1)/((x^2+x+1)) is (x^2+4x+1)/((x^2+x+1)^2)