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

The derivative of y=((x^2+1)/(x^2-1))^6 is -(24x(x^2+1)^5)/((x^2-1)^7)

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

The first derivative of y=((x^2+1)/(x^2-1))^6 is -(24x(x^2+1)^5)/((x^2-1)^7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

- \frac{24 x ( x ^{2} + 1 ) ^{5}}{( x ^{2} - 1 ) ^{7}}

Solution steps

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

Apply the chain rule:

6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} \frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 1})

= 6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} \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}}

= 6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}})

Simplify

6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}}) : - \frac{24 x ( x ^{2} + 1 ) ^{5}}{( x ^{2} - 1 ) ^{7}}

= - \frac{24 x ( x ^{2} + 1 ) ^{5}}{( x ^{2} - 1 ) ^{7}}

\frac{d}{d x} (cos (3 x) cos (2 x))

f (x) = 7 - 6 x, at x = - 1

\frac{d}{d x} (ln (tan (\frac{x}{2} + \frac{π}{4})))

\int 9 e^{3 t} d t

\frac{d}{d x} (lo g_{5} (2 x - 5))

What is the derivative of y=((x^2+1)/(x^2-1))^6 ?
The derivative of y=((x^2+1)/(x^2-1))^6 is -(24x(x^2+1)^5)/((x^2-1)^7)
What is the first derivative of y=((x^2+1)/(x^2-1))^6 ?
The first derivative of y=((x^2+1)/(x^2-1))^6 is -(24x(x^2+1)^5)/((x^2-1)^7)