Answer :
### Solution
#### Part (a)
We need to differentiate the function [tex]\( f(x) = \sqrt[3]{\frac{x+1}{x-1}} \)[/tex] with respect to [tex]\( x \)[/tex].
First, rewrite the function using exponent notation:
[tex]\[ f(x) = \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} \][/tex]
Let [tex]\( u = \frac{x+1}{x-1} \)[/tex]. Then, [tex]\( f(x) = u^{\frac{1}{3}} \)[/tex].
Now find the derivatives:
[tex]\[ \frac{d}{dx} f(x) = \frac{d}{dx} \left(u^{\frac{1}{3}}\right) = \frac{1}{3} u^{-\frac{2}{3}} \frac{du}{dx} \][/tex]
Now, find [tex]\(\frac{du}{dx}\)[/tex]:
[tex]\[ u = \frac{x+1}{x-1} \][/tex]
Using the quotient rule:
[tex]\[ \frac{du}{dx} = \frac{(x-1)\cdot 1 - (x+1)\cdot 1}{(x-1)^2} = \frac{x-1 - x - 1}{(x-1)^2} = \frac{-2}{(x-1)^2} \][/tex]
Now substitute [tex]\(\frac{du}{dx}\)[/tex] and [tex]\(u\)[/tex]:
[tex]\[ \frac{d}{dx} \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} = \frac{1}{3} \left(\frac{x+1}{x-1}\right)^{-\frac{2}{3}} \cdot \frac{-2}{(x-1)^2} \][/tex]
Simplify:
[tex]\[ \frac{d}{dx} \sqrt[3]{\frac{x+1}{x-1}} = \frac{-2}{3} \left(\frac{x+1}{x-1}\right)^{-\frac{2}{3}} \cdot \frac{1}{(x-1)^2} \][/tex]
[tex]\[ = \frac{-2}{3 (x-1)^2} \cdot \left(\frac{x-1}{x+1}\right)^{\frac{2}{3}} \][/tex]
[tex]\[ = \frac{-2}{3 (x-1)^2} \cdot \left(\frac{x-1}{x+1}\right)^{\frac{2}{3}} \][/tex]
After these steps we get:
[tex]\[ = \left(\frac{(x+1)}{(x-1)}\right)^{0.333333333333333} \cdot (x-1) \cdot (0.333333333333333/(x-1) - 0.333333333333333 \cdot (x+1)/(x-1)^2 )/(x+1) \][/tex]
#### Part (b)
We need to differentiate the function [tex]\( g(x) = \frac{\sqrt{x^2 + 1}}{(2x - 1)^2} \)[/tex] with respect to [tex]\( x \)[/tex].
First, rewrite the function using exponent notation:
[tex]\[ g(x) = \frac{(x^2 + 1)^{\frac{1}{2}}}{(2x - 1)^2} \][/tex]
We will use the quotient rule [tex]\(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)[/tex].
Let [tex]\( u = (x^2 + 1)^{\frac{1}{2}} \)[/tex] and [tex]\( v = (2x - 1)^2 \)[/tex].
First, compute their derivatives:
[tex]\[ u = (x^2 + 1)^{\frac{1}{2}} \quad \Rightarrow \quad \frac{du}{dx} = \frac{1}{2} (x^2 + 1)^{-\frac{1}{2}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \][/tex]
[tex]\[ v = (2x - 1)^2 \quad \Rightarrow \quad \frac{dv}{dx} = 2(2x - 1) \cdot 2 = 4(2x - 1) \][/tex]
Now apply the quotient rule:
[tex]\[ \frac{d}{dx} \left(\frac{\sqrt{x^2 + 1}}{(2x - 1)^2}\right) = \frac{(2x - 1)^2 \cdot \frac{x}{\sqrt{x^2 + 1}} - (x^2 + 1)^{\frac{1}{2}} \cdot 4(2x - 1)}{((2x - 1)^2)^2} \][/tex]
Simplify the numerator and denominator:
[tex]\[ = \frac{(2x - 1)^2 \cdot \frac{x}{\sqrt{x^2 + 1}} - 4(2x - 1)(x^2 + 1)^{\frac{1}{2}}}{(2x - 1)^4} \][/tex]
This simplifies to:
[tex]\[ = \frac{x}{(2x - 1)^2 \cdot \sqrt{x^2 + 1}} - \frac{4\sqrt{x^2 + 1}}{(2x - 1)^3} \][/tex]
After these steps we get:
[tex]\[ = x/((2x - 1)^2\sqrt{x^2 + 1}} - 4\sqrt{x^2 + 1}/(2x - 1)^3 \][/tex]
#### Part (c)
We need to differentiate the function [tex]\( h(x) = \frac{x^2 e^x}{(x-1)^3} \)[/tex].
Write it using quotient rule.
Let [tex]\( u = x^2 e^x \)[/tex] and [tex]\( v = (x-1)^3 \)[/tex],
First, compute these derivatives:
[tex]\[ u = x^2 e^x \quad \Rightarrow \quad \frac{du}{dx} = 2x e^x + x^2 e^x = x(x + 2) e^x \][/tex]
[tex]\[ v = (x-1)^3 \quad \Rightarrow \quad \frac{dv}{dx} = 3(x-1)^2 \][/tex]
Now apply the quotient rule:
[tex]\[ \frac{d}{dx} \left(\frac{x^2 e^x}{(x-1)^3}\right) = \frac{(x-1)^3 \cdot x(x + 2) e^x - x^2 e^x \cdot 3(x-1)^2}{(x-1)^6} \][/tex]
Simplify the numerator and denominator:
[tex]\[ = \frac{x(x+2)e^x(x-1)^3 - 3x^2 e^x (x-1)^2}{(x-1)^6} \][/tex]
This simplifies to:
[tex]\[ = \frac{e^x x^2 (x - 1)^3 - 3 e^x x^2 (x - 1)}{(x-1)^4} \][/tex]
[tex]\[ = \frac{e^x x^2 ((x - 1)(x+2) - 3(x+1))}{(x-1)^4} \][/tex]
This simplifies to:
[tex]\[ = \frac{e^x x^2 (x(x - 1)+x(x + 2)- 3)}{(x-1)^4} \][/tex]
This can be written step by step as the following:
[tex]\[ = x^2 e^x / (x-1)^3 -3 x^2 e(x)/ (x-1)^4 + 2 x e(x)/(x-1)^3 \][/tex]
Therefore:
[tex]\[ = x^2 exp(x)/(x - 1)^3 - 3 x^2 exp(x)/(x - 1)^4 + 2 x * exp(x)/(x -1)^3 \][/tex]
#### Part (a)
We need to differentiate the function [tex]\( f(x) = \sqrt[3]{\frac{x+1}{x-1}} \)[/tex] with respect to [tex]\( x \)[/tex].
First, rewrite the function using exponent notation:
[tex]\[ f(x) = \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} \][/tex]
Let [tex]\( u = \frac{x+1}{x-1} \)[/tex]. Then, [tex]\( f(x) = u^{\frac{1}{3}} \)[/tex].
Now find the derivatives:
[tex]\[ \frac{d}{dx} f(x) = \frac{d}{dx} \left(u^{\frac{1}{3}}\right) = \frac{1}{3} u^{-\frac{2}{3}} \frac{du}{dx} \][/tex]
Now, find [tex]\(\frac{du}{dx}\)[/tex]:
[tex]\[ u = \frac{x+1}{x-1} \][/tex]
Using the quotient rule:
[tex]\[ \frac{du}{dx} = \frac{(x-1)\cdot 1 - (x+1)\cdot 1}{(x-1)^2} = \frac{x-1 - x - 1}{(x-1)^2} = \frac{-2}{(x-1)^2} \][/tex]
Now substitute [tex]\(\frac{du}{dx}\)[/tex] and [tex]\(u\)[/tex]:
[tex]\[ \frac{d}{dx} \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} = \frac{1}{3} \left(\frac{x+1}{x-1}\right)^{-\frac{2}{3}} \cdot \frac{-2}{(x-1)^2} \][/tex]
Simplify:
[tex]\[ \frac{d}{dx} \sqrt[3]{\frac{x+1}{x-1}} = \frac{-2}{3} \left(\frac{x+1}{x-1}\right)^{-\frac{2}{3}} \cdot \frac{1}{(x-1)^2} \][/tex]
[tex]\[ = \frac{-2}{3 (x-1)^2} \cdot \left(\frac{x-1}{x+1}\right)^{\frac{2}{3}} \][/tex]
[tex]\[ = \frac{-2}{3 (x-1)^2} \cdot \left(\frac{x-1}{x+1}\right)^{\frac{2}{3}} \][/tex]
After these steps we get:
[tex]\[ = \left(\frac{(x+1)}{(x-1)}\right)^{0.333333333333333} \cdot (x-1) \cdot (0.333333333333333/(x-1) - 0.333333333333333 \cdot (x+1)/(x-1)^2 )/(x+1) \][/tex]
#### Part (b)
We need to differentiate the function [tex]\( g(x) = \frac{\sqrt{x^2 + 1}}{(2x - 1)^2} \)[/tex] with respect to [tex]\( x \)[/tex].
First, rewrite the function using exponent notation:
[tex]\[ g(x) = \frac{(x^2 + 1)^{\frac{1}{2}}}{(2x - 1)^2} \][/tex]
We will use the quotient rule [tex]\(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)[/tex].
Let [tex]\( u = (x^2 + 1)^{\frac{1}{2}} \)[/tex] and [tex]\( v = (2x - 1)^2 \)[/tex].
First, compute their derivatives:
[tex]\[ u = (x^2 + 1)^{\frac{1}{2}} \quad \Rightarrow \quad \frac{du}{dx} = \frac{1}{2} (x^2 + 1)^{-\frac{1}{2}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \][/tex]
[tex]\[ v = (2x - 1)^2 \quad \Rightarrow \quad \frac{dv}{dx} = 2(2x - 1) \cdot 2 = 4(2x - 1) \][/tex]
Now apply the quotient rule:
[tex]\[ \frac{d}{dx} \left(\frac{\sqrt{x^2 + 1}}{(2x - 1)^2}\right) = \frac{(2x - 1)^2 \cdot \frac{x}{\sqrt{x^2 + 1}} - (x^2 + 1)^{\frac{1}{2}} \cdot 4(2x - 1)}{((2x - 1)^2)^2} \][/tex]
Simplify the numerator and denominator:
[tex]\[ = \frac{(2x - 1)^2 \cdot \frac{x}{\sqrt{x^2 + 1}} - 4(2x - 1)(x^2 + 1)^{\frac{1}{2}}}{(2x - 1)^4} \][/tex]
This simplifies to:
[tex]\[ = \frac{x}{(2x - 1)^2 \cdot \sqrt{x^2 + 1}} - \frac{4\sqrt{x^2 + 1}}{(2x - 1)^3} \][/tex]
After these steps we get:
[tex]\[ = x/((2x - 1)^2\sqrt{x^2 + 1}} - 4\sqrt{x^2 + 1}/(2x - 1)^3 \][/tex]
#### Part (c)
We need to differentiate the function [tex]\( h(x) = \frac{x^2 e^x}{(x-1)^3} \)[/tex].
Write it using quotient rule.
Let [tex]\( u = x^2 e^x \)[/tex] and [tex]\( v = (x-1)^3 \)[/tex],
First, compute these derivatives:
[tex]\[ u = x^2 e^x \quad \Rightarrow \quad \frac{du}{dx} = 2x e^x + x^2 e^x = x(x + 2) e^x \][/tex]
[tex]\[ v = (x-1)^3 \quad \Rightarrow \quad \frac{dv}{dx} = 3(x-1)^2 \][/tex]
Now apply the quotient rule:
[tex]\[ \frac{d}{dx} \left(\frac{x^2 e^x}{(x-1)^3}\right) = \frac{(x-1)^3 \cdot x(x + 2) e^x - x^2 e^x \cdot 3(x-1)^2}{(x-1)^6} \][/tex]
Simplify the numerator and denominator:
[tex]\[ = \frac{x(x+2)e^x(x-1)^3 - 3x^2 e^x (x-1)^2}{(x-1)^6} \][/tex]
This simplifies to:
[tex]\[ = \frac{e^x x^2 (x - 1)^3 - 3 e^x x^2 (x - 1)}{(x-1)^4} \][/tex]
[tex]\[ = \frac{e^x x^2 ((x - 1)(x+2) - 3(x+1))}{(x-1)^4} \][/tex]
This simplifies to:
[tex]\[ = \frac{e^x x^2 (x(x - 1)+x(x + 2)- 3)}{(x-1)^4} \][/tex]
This can be written step by step as the following:
[tex]\[ = x^2 e^x / (x-1)^3 -3 x^2 e(x)/ (x-1)^4 + 2 x e(x)/(x-1)^3 \][/tex]
Therefore:
[tex]\[ = x^2 exp(x)/(x - 1)^3 - 3 x^2 exp(x)/(x - 1)^4 + 2 x * exp(x)/(x -1)^3 \][/tex]
