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

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Answer 1 · 2016-07-22T12:49:42

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

I'm assuming you mean differentiate? I don't think "derive" is the correct word in this context. The quotient rule is given 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)#

Tidying up and removing negative exponents:

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