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

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

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

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

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derivative of

g (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}}

x \to 0 lim (2 x - p (x))

\frac{d}{d x} (\frac{π x}{12})

6 \frac{d y}{d x} + 3 e^{x + y} = 0

g (x) = x, (1, 1)

\frac{\partial}{\partial x} (3 x cos (4 x y))

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