How do you differentiate v=(sqrtx+1/root3x)^2?

1 Answer
Apr 19, 2018

v'(x) = 1 + 1/3x^(-5/6) -2/3x^(-5/3)

Explanation:

write both x-terms as numerical exponents of x:

sqrt(x) = x^(1/2)

root3(x) = x^(1/3)

1/(root3(x)) = x^(-1/3)

(sqrt(x) + 1/(root3(x)))^2 = (x^(1/2) + x^(-1/3))^2

using FOIL:

(x^(1/2) + x^(-1/3))(x^(1/2) + x^(-1/3)) = x^1 + x^(1/6) + x^(1/6) + x^(-2/3)

then you can differentiate these values separately, using the power rule

f(x) = nx^(n-1).

when x^n = x^1, nx^(n-1) is x^0, or 1.

when x^n = x^(1/6), nx^(n-1) is 1/6x^(-5/6).

when x^n = x^(-2/3), nx^(n-1) is -2/3x^(-5/3).

hence, when x^1 + x^(1/6) + x^(1/6) + x^(-2/3) is differentiated, the result is

1 + 1/6x^(-5/6) + 1/6x^(-5/6) + -2/3x^(-5/3).

this can be simplified to 1 + 1/3x^(-5/6) -2/3x^(-5/3).

or even (3 + x^(-5/6) - 2x^(-5/3))/3.

v'(x) = 1 + 1/3x^(-5/6) -2/3x^(-5/3).