What is the derivative of h(x)=[(x+1)/(x-1)]^3 ?

The derivative of h(x)=[(x+1)/(x-1)]^3 is -(6(x+1)^2)/((x-1)^4)

What is the first derivative of h(x)=[(x+1)/(x-1)]^3 ?

The first derivative of h(x)=[(x+1)/(x-1)]^3 is -(6(x+1)^2)/((x-1)^4)

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derivative of

h (x) = [\frac{x + 1}{x - 1}]^{3}

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} ((\frac{y}{z})^{\frac{1}{x}})

y^{^{''}} + 3 y = 48 x^{2} e^{3 x}

d er i v a t i v e f (x) = x^{- 4}

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

d er i v a t i v e x^{3} sin (7 x)

What is the derivative of h(x)=[(x+1)/(x-1)]^3 ?
The derivative of h(x)=[(x+1)/(x-1)]^3 is -(6(x+1)^2)/((x-1)^4)
What is the first derivative of h(x)=[(x+1)/(x-1)]^3 ?
The first derivative of h(x)=[(x+1)/(x-1)]^3 is -(6(x+1)^2)/((x-1)^4)