Question

How do you find the derivative of $y=\sqrt{2x}$ ? $$$$

Answer 1

We recall the definition of composite function $gof\left( x \right)=g\left( f\left( x \right) \right)$. We recall the chain rule of differentiation $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$ where $y=gof=\sqrt{2x}$ and $u=f\left( x \right)=2x$. We first find $u=f\left( x \right)$ as the function inside the square root and $y$ as the given function and then differentiate using chain rule. $$$$

Complete step by step answer:
If the functions $f\left( x \right),g\left( x \right)$ are defined within sets $f:A\to B$ and $g:B\to C$ then the composite function from A to C is defend as $g\left( f\left( x \right) \right)$ within sets $gof:A\to C$. If we denote $g\left( f\left( x \right) \right)=y$ and $f\left( x \right)=u$ then we can differentiate the composite function using chain rule as
$\dfrac{d}{dx}g\left( f\left( x \right) \right)=\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$
We are asked to differentiate the function ${{\left( \sqrt{2x} \right)}^{n}}$. We see that it is a composite function made by functions polynomial square root function that is ${{x}^{n}}$ and polynomial function that is $2x$. Let us assign the function within the bracket as $f\left( x \right)=2x=u$ and $g\left( x \right)=\sqrt{x}$. So we have$g\left( f\left( x \right) \right)=g\left( 2x \right)=\sqrt{2x}=y$. We differentiate using chain rule to have;

& \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \\\ & \Rightarrow \dfrac{d}{dx}y=\dfrac{d}{du}y\times \dfrac{d}{dx}u \\\ & \Rightarrow \dfrac{d}{dx}\sqrt{2x}=\dfrac{d}{d\left( 2x \right)}\sqrt{2x}\times \dfrac{d}{dx}\left( 2x \right) \\\ \end{aligned}$$ We use $\sqrt{a}={{a}^{\dfrac{1}{2}}}$ to have $$\Rightarrow \dfrac{d}{dx}\sqrt{2x}=\dfrac{d}{d\left( 2x \right)}{{\left( 2x \right)}^{\dfrac{1}{2}}}\times \dfrac{d}{dx}\left( 2x \right)$$ We know that from standard differentiation of polynomial function as $\dfrac{d}{dt}{{t}^{n}}=n{{t}^{n-1}}$ where $n$ is any real number. We use it for $t=2x,n=\dfrac{1}{2}$ in the above step to have $$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\sqrt{2x}=\dfrac{1}{2}{{\left( 2x \right)}^{\dfrac{1}{2}-1}}\dfrac{d}{dx}\left( 2x \right) \\\ & \Rightarrow \dfrac{d}{dx}\sqrt{2x}=\dfrac{1}{2}{{\left( 2x \right)}^{-\dfrac{1}{2}}}\dfrac{d}{dx}\left( 2x \right) \\\ \end{aligned}$$ We take out the constant coefficient $2$ of $2x$ outside the differentiation using $\dfrac{d}{dx}cf\left( x \right)=c\dfrac{d}{x}f\left( x \right)$ to have $$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\sqrt{2x}=\dfrac{1}{2}{{\left( 2x \right)}^{-\dfrac{1}{2}}}\times 2\dfrac{d}{dx}\left( x \right) \\\ & \Rightarrow \dfrac{d}{dx}\sqrt{2x}=\dfrac{1}{2}\times 2\times {{\left( 2x \right)}^{-\dfrac{1}{2}}}\times 1 \\\ & \Rightarrow \dfrac{d}{dx}\sqrt{2x}=1\times \dfrac{1}{\sqrt{2x}}\times 1 \\\ & \Rightarrow \dfrac{d}{dx}\sqrt{2x}=\dfrac{1}{\sqrt{2x}} \\\ \end{aligned}$$ **Note:** We can alternatively solve by taking out the 2 outside the square root as $$\begin{aligned} & y=\sqrt{2x} \\\ & \Rightarrow y=\sqrt{2}\cdot \sqrt{x} \\\ \end{aligned}$$ We differentiate both sides of the above step with respect to $x$. We take the constant $\sqrt{2}$ outside the square root to have; $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{2}\cdot \sqrt{x} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\sqrt{2}\dfrac{d}{dx}\left( \sqrt{x} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\sqrt{2}\dfrac{d}{dx}\left( {{x}^{\dfrac{1}{2}}} \right) \\\ \end{aligned}$$ We use standard differentiation of polynomial function as $\dfrac{d}{dt}{{t}^{n}}=n{{t}^{n-1}}$ for $t=x,n=\dfrac{1}{2}$ in the above step to have $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\sqrt{2}\cdot \dfrac{1}{2}\cdot {{x}^{\dfrac{1}{2}-1}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{2}}\cdot {{x}^{-\dfrac{1}{2}}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{2x}} \\\ \end{aligned}$$ We should remember $\dfrac{d}{dx}\sqrt{f\left( x \right)}=\dfrac{1}{\sqrt{f\left( x \right)}}$ for all $f\left( x \right)>0$. We note the square function maps a non-negative number to a non-negative number. So here $2x\ge 0$ but $x=0$ the $f\left( x \right)=\sqrt{2x}$ is not differentiable.