How do you derive #y = (x-1)/( x+3) ^ (1/3)# using the quotient rule?

Question

Christopher Allers · Accepted Answer

#(dy)/(dx) = 1/(root(3)(x+3)) - (x-1)/(3(x+3)^(4/3))#

I take it that you mean to differentiate? "Derive" isn't the right word in this situation. The quotient rule is provided by:

#d/(dx)((f(x))/(g(x))) = (f'(x)g(x) - f(x)g'(x))/(g(x))^2#

We have #f(x) = x-1 implies f'(x) = 1#

and

#g(x) = (x+3)^(1/3) implies g'(x) = 1/3(x+3)^(-2/3)#

Hence #(dy)/(dx) = (1*(x+3)^(1/3) - (x-1)*1/3(x+3)^(-2/3))/((x+3)^(1/3))^2#

#(dy)/(dx) = ((x+3)^(1/3) - (x-1)*1/3(x+3)^(-2/3))/((x+3)^(2/3))#

#(dy)/(dx) = (x+3)^(-1/3) - (x-1)/3(x+3)^(-4/3)#

Organizing and eliminating negative exponents:

#(dy)/(dx) = 1/(root(3)(x+3)) - (x-1)/(3(x+3)^(4/3))#

Christopher Agresta · Answer

undefined To derive $ y = \frac{{x - 1}}{{(x + 3)^{\frac{1}{3}}}} $ using the quotient rule, we differentiate the numerator and denominator separately, and then apply the quotient rule formula.

Given:
$$ y = \frac{{u}}{{v}} $$
Where $ u = x - 1 $ and $ v = (x + 3)^{\frac{1}{3}} $.

The quotient rule states:
$$ \frac{{dy}}{{dx}} = \frac{{v \cdot \frac{{du}}{{dx}} - u \cdot \frac{{dv}}{{dx}}}}{{v^2}} $$

First, let's find $ \frac{{du}}{{dx}} $ and $ \frac{{dv}}{{dx}} $:
$$ \frac{{du}}{{dx}} = \frac{{d}}{{dx}}(x - 1) = 1 $$
$$ \frac{{dv}}{{dx}} = \frac{{d}}{{dx}}((x + 3)^{\frac{1}{3}}) = \frac{1}{3}(x + 3)^{-\frac{2}{3}} \cdot 1 = \frac{1}{3(x + 3)^{\frac{2}{3}}} $$

Now, substitute $ u $, $ v $, $ \frac{{du}}{{dx}} $, and $ \frac{{dv}}{{dx}} $ into the quotient rule formula:
$$ \frac{{dy}}{{dx}} = \frac{{(x + 3)^{\frac{1}{3}} \cdot 1 - (x - 1) \cdot \frac{1}{3(x + 3)^{\frac{2}{3}}}}}{{(x + 3)^{\frac{1}{3}})^2}} $$

Simplify the expression:
$$ \frac{{(x + 3)^{\frac{1}{3}} - \frac{{x - 1}}{{3(x + 3)^{\frac{2}{3}}}}}}{{(x + 3)^{\frac{2}{3}}}} $$

Multiply both the numerator and the denominator by $ (x + 3)^{\frac{1}{3}} $ to eliminate the fractional exponents in the denominator:
$$ \frac{{(x + 3) - \frac{{x - 1}}{{3}}}}{{(x + 3)}} \cdot \frac{{(x + 3)^{\frac{1}{3}}}}{{(x + 3)^{\frac{1}{3}}}} $$

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

$$ \frac{{3x + 9 - x + 1}}{{3(x + 3)^{\frac{1}{3}} \cdot (x + 3)}} $$

$$ \frac{{2x + 10}}{{3(x + 3)^{\frac{1}{3}} \cdot (x + 3)}} $$

$$ \frac{{2(x + 5)}}{{3(x + 3)^{\frac{1}{3}} \cdot (x + 3)}} $$

Therefore, $ \frac{{dy}}{{dx}} = \frac{{2(x + 5)}}{{3(x + 3)^{\frac{1}{3}} \cdot (x + 3)}} $.