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}}

\frac{d}{d x} (\frac{1}{3} ln (x))

7 ln (y) y^{^{'}} - x^{9} y = 0

\int sin (2 x) \cdot sec (x) d x

\int \frac{6 x ^{3} - 12 x ^{2} + 4}{x ^{2} - 2 x} d x

x \to 2 π lim (\frac{x - 2 π}{sin ( 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)