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

a) $dy/dx=3/2\sqrt x$ b) $dy/dx=3/2\sqrt x$
a) $y=u^3$ and $u=\sqrt x=x^{1/2}$ We have $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(u^3)'(x^{1/2})'=3u^2\times\frac{1}{2}x^{-1/2}=\frac{3}{2}u^2x^{-1/2}$$ Since $u=x^{1/2}$, here we substitute $u^2=(x^{1/2})^2=x$: $$\frac{dy}{dx}=\frac{3}{2}x\times x^{-1/2}=\frac{3}{2}x^{1/2}=\frac{3}{2}\sqrt x$$ b) $y=\sqrt u=u^{1/2}$ and $u=x^3$ We have $$\frac{dy}{dx}=(u^{1/2})'(x^3)'=\frac{1}{2}u^{-1/2}\times3x^2=\frac{3}{2}u^{-1/2}x^2$$ Since $u=x^3$, here we substitute $u^{-1/2}=(x^3)^{-1/2}=x^{-3/2}$: $$\frac{dy}{dx}=\frac{3}{2}x^{-3/2}x^2=\frac{3}{2}x^{1/2}=\frac{3}{2}\sqrt x$$ For $y=x^{3/2}$, we have $dy/dx=3/2x^{1/2}=3/2\sqrt x$ So in both cases, we see that the results are the same and equal with $dy/dx$ when $y=x^{3/2}$. That means writing the function as a composite in different ways does not change the value of its derivative.