## University Calculus: Early Transcendentals (3rd Edition)

$$x^{1/4}=\sqrt{\sqrt x}=(x^{1/2})^{1/2}$$ Using the Chain Rule, we have $$\frac{d}{dx}(x^{1/2})^{1/2}=\frac{1}{2}(x^{1/2})^{-1/2}\frac{d}{dx}(x^{1/2})=\frac{1}{2}(x^{1/2})^{-1/2}\times\frac{1}{2}x^{-1/2}$$ $$=\frac{1}{4}x^{-1/4}\times x^{-1/2}=\frac{1}{4}x^{-3/4}=\frac{1}{4}x^{1/4-1}$$ So the Power Rule $(x^n)'=nx^{n-1}$ holds for the function $x^{1/4}$.