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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\frac{d}{d x} (ln (sin (x)) + cos^{2} (x))

\int \frac{y ^{2} - 81}{y} d y

n = 0 \sum \infty \frac{1}{2 ^{n} + 3 ^{n}}

\int cos (2 n x) d x

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