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

The derivative of y=((x^2+1)/(x^2-1))^3 is -(12x(x^2+1)^2)/((x^2-1)^4)

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

The first derivative of y=((x^2+1)/(x^2-1))^3 is -(12x(x^2+1)^2)/((x^2-1)^4)

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derivative of

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

- \frac{12 x ( x ^{2} + 1 ) ^{2}}{( x ^{2} - 1 ) ^{4}}

Solution steps

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

Apply the chain rule:

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

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

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

= 3 (\frac{x ^{2} + 1}{x ^{2} - 1})^{2} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}})

Simplify

3 (\frac{x ^{2} + 1}{x ^{2} - 1})^{2} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}}) : - \frac{12 x ( x ^{2} + 1 ) ^{2}}{( x ^{2} - 1 ) ^{4}}

= - \frac{12 x ( x ^{2} + 1 ) ^{2}}{( x ^{2} - 1 ) ^{4}}

\int_{1}^{4} 5 - 2 t + 3 t^{2} d t

\frac{\partial}{\partial x} (2 x + 3 y + 4)

\int 2 x (x^{2} + 5)^{4} d x

f (x) = x^{2} (4 - x)^{2}

9 x y \frac{d y}{d x} = 1

What is the derivative of y=((x^2+1)/(x^2-1))^3 ?
The derivative of y=((x^2+1)/(x^2-1))^3 is -(12x(x^2+1)^2)/((x^2-1)^4)
What is the first derivative of y=((x^2+1)/(x^2-1))^3 ?
The first derivative of y=((x^2+1)/(x^2-1))^3 is -(12x(x^2+1)^2)/((x^2-1)^4)