What is the derivative of (ax-4sqrt(1+x)/(x^2-1)) ?

The derivative of (ax-4sqrt(1+x)/(x^2-1)) is (-ax^2+6xsqrt(x+1)+2sqrt(x+1)-a)/((x^2-1)^2)

What is the first derivative of (ax-4sqrt(1+x)/(x^2-1)) ?

The first derivative of (ax-4sqrt(1+x)/(x^2-1)) is (-ax^2+6xsqrt(x+1)+2sqrt(x+1)-a)/((x^2-1)^2)

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

\frac{- a x ^{2} + 6 x x + 1 + 2 x + 1 - a}{( x ^{2} - 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(a - \frac{2}{1 + x}) (x^{2} - 1) - 2 x (a x - 4 1 + x) : - a x^{2} + 6 x x + 1 + 2 x + 1 - a

= \frac{- a x ^{2} + 6 x x + 1 + 2 x + 1 - a}{( x ^{2} - 1 ) ^{2}}

n = 1 \sum \infty (\frac{1}{e})^{n}

\int \frac{2}{3} x d x

\frac{\partial}{\partial x} (1 + x^{2} + y^{2})

\int_{1 - 2 x}^{1 + 2 x} t sin (t) d t

What is the derivative of (ax-4sqrt(1+x)/(x^2-1)) ?
The derivative of (ax-4sqrt(1+x)/(x^2-1)) is (-ax^2+6xsqrt(x+1)+2sqrt(x+1)-a)/((x^2-1)^2)
What is the first derivative of (ax-4sqrt(1+x)/(x^2-1)) ?
The first derivative of (ax-4sqrt(1+x)/(x^2-1)) is (-ax^2+6xsqrt(x+1)+2sqrt(x+1)-a)/((x^2-1)^2)