Derivative of a chain of three functions

Text

Let $z=2x^{3}$, $v=\dfrac{5}{z^{2}}$ and $y=\sqrt{1+v}$. Find $\dfrac{dy}{dx}$ as a function of $x$.

Solution

Three-factor chain rule $\dfrac{dy}{dx}=\dfrac{dy}{dv}\cdot\dfrac{dv}{dz}\cdot\dfrac{dz}{dx}$. $\frac{dy}{dv}=\frac{1}{2\sqrt{1+v}},\qquad \frac{dv}{dz}=-\frac{10}{z^{3}},\qquad \frac{dz}{dx}=6x^{2}.$ With $z=2x^{3}$ we get $z^{3}=8x^{9}$, so $\dfrac{dv}{dz}=-\dfrac{5}{4x^{9}}$ and $\frac{dy}{dx}=\frac{1}{2\sqrt{1+v}}\cdot\left(-\frac{5}{4x^{9}}\right)\cdot 6x^{2}=-\frac{15}{4x^{7}\sqrt{1+v}}.$ Since $v=\dfrac{5}{4x^{6}}$, we have $\sqrt{1+v}=\dfrac{\sqrt{4x^{6}+5}}{2x^{3}}$, hence $\frac{dy}{dx}=-\frac{15}{2x^{4}\sqrt{4x^{6}+5}}.$ $\boxed{\dfrac{dy}{dx}=-\dfrac{15}{2x^{4}\sqrt{4x^{6}+5}}}$