Answer to Question #340775 in Calculus for Jayson

Question

Solve using basic differentiation rule

y = 2/(x ^ (1/2)) + 6/(x ^ (1/3)) - 2/(x ^ (3/2)) + 4/(x ^ (3/4))

Answer 1

$y=\frac{2}{x^{1/2}}+\frac{6}{x^{1/3}}-\frac{2}{x^{3/2}}+\frac{4}{x^{3/4}}=2x^{-1/2}+6x^{-1/3}-2x^{-3/2}+4x^{-3/4}$

$y'=-x^{-3/2}-2x^{-4/3}+3x^{-5/2}-3x^{-7/4}=-\frac{1}{x^{3/2}}-\frac{2}{x^{4/3}}+\frac{3}{x^{5/2}}-\frac{3}{x^{7/4}}$