What is the derivative of y= 3/(x^2-1) ?

The derivative of y= 3/(x^2-1) is -(6x)/((x^2-1)^2)

What is the first derivative of y= 3/(x^2-1) ?

The first derivative of y= 3/(x^2-1) is -(6x)/((x^2-1)^2)

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derivative of

y = \frac{3}{x ^{2} - 1}

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

Solution steps

\frac{d}{d x} (\frac{3}{x ^{2} - 1})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} (\frac{1}{x ^{2} - 1})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 3 \frac{d}{d x} ((x^{2} - 1)^{- 1})

Apply the chain rule:

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

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

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

= 3 (- \frac{1}{( x ^{2} - 1 ) ^{2}} \cdot 2 x)

Simplify

3 (- \frac{1}{( x ^{2} - 1 ) ^{2}} \cdot 2 x) : - \frac{6 x}{( x ^{2} - 1 ) ^{2}}

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

x \to 0 + lim (6 (cot (x) - \frac{1}{x}))

\frac{d}{d x} (- e^{- 2 x})

f (x) = 5 x + 4, at x = 9

y^{^{''}} + 10 y^{^{'}} + 25 y = 2 r^{- 5 x}

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

What is the derivative of y= 3/(x^2-1) ?
The derivative of y= 3/(x^2-1) is -(6x)/((x^2-1)^2)
What is the first derivative of y= 3/(x^2-1) ?
The first derivative of y= 3/(x^2-1) is -(6x)/((x^2-1)^2)