How do you differentiate #f(x)=x/(x-1)^2+x^2-4/(1-1x)# using the sum rule?

Question

Charles Anctil · Accepted Answer

#f'(x)=(2x)/(x-1)^3+(x-1)^2+2x-4/(1-x)^2# #"differentiate using the "color(blue)"chain/product rules"# #"given "y=f(g(x))"then"# #dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"# #"given "y=g(x)h(x)" then"# #dy/dx=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"# #"differentiate "x/(x-1)^2" using chain/product rules"# #x/(x-1)^2=x(x-1)^-2# #d/dx(x(x-1)^2)# #=x(-2(x-1)^-3)+(x-1)^2# #=-2x(x-1)^-3+(x-1)^2=(2x)/(x-1)^3+(x-1)^2# #"differentiate "4/(1-x)" using the chain rule"# #4/(1-x)=4(1-x)^-1# #d/dx(4(1-x)^-1)=-4(1-x)^-2(-1)=4/(1-x)^2# #"returning to the original question"# #f'(x)=(2x)/(x-1)^3+(x-1)^2+2x-4/(1-x)^2#

Luke Alvarez · Answer

undefined To differentiate $ f(x) = \frac{x}{(x - 1)^2} + \frac{x^2 - 4}{1 - x} $ using the sum rule:

1. Differentiate each term separately.
2. Apply the sum rule to combine the derivatives.

Given $ f(x) = \frac{x}{(x - 1)^2} + \frac{x^2 - 4}{1 - x} $, let's differentiate each term:

For the first term, $ \frac{x}{(x - 1)^2} $:
- Apply the quotient rule.
- $ u = x $ and $ v = (x - 1)^2 $.
- Differentiate $ u $ and $ v $ separately.

$$ \frac{d}{dx} \left( \frac{x}{(x - 1)^2} \right) = \frac{u'v - uv'}{v^2} $$

For the second term, $ \frac{x^2 - 4}{1 - x} $:
- Apply the quotient rule.
- $ u = x^2 - 4 $ and $ v = 1 - x $.
- Differentiate $ u $ and $ v $ separately.

$$ \frac{d}{dx} \left( \frac{x^2 - 4}{1 - x} \right) = \frac{u'v - uv'}{v^2} $$

Once you have the derivatives of each term, apply the sum rule:

$$ f'(x) = \frac{d}{dx} \left( \frac{x}{(x - 1)^2} \right) + \frac{d}{dx} \left( \frac{x^2 - 4}{1 - x} \right) $$

Finally, simplify the resulting expression to obtain $ f'(x) $.