How do you find the derivative for #y = ((x-1)/( x+3)) ^ (1/3)#?

Question

Samantha Adkison · Accepted Answer

You can use the quotient rule and the chain rule.

You can differentiate your function by using the chain rule, the quotient rule, and the power rule. First, you need to recognize that your function can be expressed as #f(u) = u^(1/3)#, where #u = (x-1)/(x+3)# The chain rule allows you to differentiate a function #y# that depends on a variable #u#, which in turn depends on a variable #x# by using the formula #color(blue)(d/dx(y) = d/(du)(y) * d/dx(u)# The derivative of #y# will thus be equal to #d/dx(y) = d/(du)(u^(1/3)) * d/dx((x-1)/(x-3))# #d/dx(y) = 1/3u^(1/3 - 1) * d/dx((x-1)/(x+3))# To determine #d/dx((x-1)/(x+3))#, you can use the quotient rule, which allows you to differentiate functions that can be expressed as a quotient of two other functions #a(x) = (b(x))/(c(x))# by using the formula #color(blue)(d/dx(a(x)) = (b^'(x) * c(x) - b(x) * c^'(x))/[c(x)]^2#, with #c(x)!=0#. In your case, #d/dx((x-1)/(x+3))# can be differentiated by using the quotient rule to get #d/dx((x-1)/(x+3)) = ([d/dx(x-1)] * (x+3) - (x-1) * d/dx(x+3))/(x+3)^2# #d/dx((x-1)/(x+3)) = (color(red)(cancel(color(black)(x))) + 3 - color(red)(cancel(color(black)(x))) + 1)/(x+3)^2# #d/dx((x-1)/(x+3)) = 4/(x+3)^2# Plug this into the derivative of #y# to get #d/dx(y) = 1/3 * ((x-1)/(x+3))^(-2/3) * 4/(x+3)^2# #d/dx(y) = 4/3 * 1/( (x-1)^(2/3) * 1/(x+3)^(2/3) * (x+3)^2)# #d/dx(y) = 4/3 * 1/((x-1)^(2/3) * (x+3)^(4/3)) = color(green)(4/3 * (x-1)^(-2/3) * (x+3)^(-4/3))# Alternatively, you can complicate things a bit by writing your original function as #y = (x-1)^(1/3)/(x+3)^(1/3)# and using a combination of the power rule and quotient rule. This will get a little messy, but you could write #d/dx(y) = ([d/dx(x-1)^(1/3)] * (x + 3)^(1/3) - (x-1)^(1/3) * d/dx(x+3)^(1/3))/[(x + 3)^(1/3)]^2# #d/dx(y) = (1/3 (x-1)^(-2/3) * (x+3)^(1/3) - (x-1)^(1/3) * 1/3 * (x+3)^(-2/3))/(x+3)^(2/3)# #d/dx(y) = 1/3 * ((x-1)^(-2/3) * (x + 3)^(1/3) - (x-1)^(1/3) * (x+3)^(-2/3))/(x + 3)^(2/3)# This is equivalent to #d/dx(y) = 1/3 * [(x-1)^(-2/3) * (x+3)^(-1/3) - (x-1)^(1/3) * (x + 3)^(-4/3)]# #d/dx(y) = 1/3 * [(x-1)^(1/3) * (x + 3)^(-1/3) [(x-1)^(-1) - (x + 3)^(-1)]]# #d/dx(y) = 1/3 * [(x-1)^(1/3) * (x + 3)^(-1/3) * (color(red)(cancel(color(black)(x))) + 3 - color(red)(cancel(color(black)(x))) + 1)/((x-1)(x+3))]# Finally, you have #d/dx(y) = color(green)(4/3 * (x-1)^(-2/3) * (x+3)^(-4/3))#

Ezekiel Alexander · Answer

undefined To find the derivative of $ y = \left(\frac{x-1}{x+3}\right)^{\frac{1}{3}} $, we can use the chain rule. The derivative is given by:

$$ y' = \frac{1}{3} \left(\frac{x-1}{x+3}\right)^{\frac{1}{3}-1} \times \frac{d}{dx} \left(\frac{x-1}{x+3}\right) $$

$$ = \frac{1}{3} \left(\frac{x-1}{x+3}\right)^{-\frac{2}{3}} \times \left( \frac{(x+3)(1) - (x-1)(1)}{(x+3)^2} \right) $$

$$ = \frac{1}{3} \left(\frac{x-1}{x+3}\right)^{-\frac{2}{3}} \times \frac{4}{(x+3)^2} $$

$$ = \frac{4}{3(x-1)^{2/3}(x+3)^{5/3}} $$