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

The derivative of ((x^2+1/(x^2-1))^4) is -(16x(x^2+1)^3)/((x^2-1)^5)

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

The first derivative of ((x^2+1/(x^2-1))^4) is -(16x(x^2+1)^3)/((x^2-1)^5)

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

- \frac{16 x ( x ^{2} + 1 ) ^{3}}{( x ^{2} - 1 ) ^{5}}

Solution steps

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

Apply the chain rule:

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

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

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

Simplify

4 (\frac{x ^{2} + 1}{x ^{2} - 1})^{3} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}}) : - \frac{16 x ( x ^{2} + 1 ) ^{3}}{( x ^{2} - 1 ) ^{5}}

= - \frac{16 x ( x ^{2} + 1 ) ^{3}}{( x ^{2} - 1 ) ^{5}}

y = cos (\frac{π t}{2} - \frac{π}{3})

h \to 0 lim (\frac{1 - cos ( 7 h )}{h})

- 2 x^{2} + 2 x

x \to 6 lim (21 x + 18)

x \to - \infty lim (- (x))

What is the derivative of ((x^2+1/(x^2-1))^4) ?
The derivative of ((x^2+1/(x^2-1))^4) is -(16x(x^2+1)^3)/((x^2-1)^5)
What is the first derivative of ((x^2+1/(x^2-1))^4) ?
The first derivative of ((x^2+1/(x^2-1))^4) is -(16x(x^2+1)^3)/((x^2-1)^5)