Chapter 3, Problem 4RQ

To determine

Whether the statement, “If f is differentiable, then $\frac{d}{dx}\left[\sqrt{f\left(x\right)}\right]=\frac{f^{\prime }\left(x\right)}{2\sqrt{f\left(x\right)}}$” is true or false.

Answer to Problem 4RQ

The statement is true.

Explanation of Solution

Result used: Chain Rule

If g is differentiable at t and f is differentiable at $g\left(t\right)$, then the composite function $F=f\circ g$ defined by $F\left(t\right)=f\left(g\left(t\right)\right)$ is differentiable at x and $F^{\prime }$ is given by the product

$F^{\prime }\left(t\right)=f^{\prime }\left(g\left(t\right)\right)\cdot g^{\prime }\left(t\right)$ (1)

Calculation:

Consider $\frac{d}{dx}\left[\sqrt{f\left(x\right)}\right]=\frac{f^{\prime }\left(x\right)}{2\sqrt{f\left(x\right)}}$.

Take the left hand side $\frac{d}{dx}\left[\sqrt{f\left(x\right)}\right]$.

$\frac{d}{dx}\left[\sqrt{f\left(x\right)}\right]=\frac{d}{dx}{\left[f\left(x\right)\right]}^{\frac{1}{2}}$

Let $h\left(x\right)=f\left(x\right)$ and $g\left(u\right)=u^{\frac{1}{2}}\text{ where }u=h\left(x\right)$.

Apply the chain rule as shown in equation (1),

$\frac{d}{dx}\left[\sqrt{f\left(x\right)}\right]=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (2)

The derivative $f^{\prime }\left(h\left(x\right)\right)$ is computed as follows,

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)=g^{\prime }\left(u\right)\left[Qu=h\left(x\right)\right]\\ =\frac{d}{du}\left(g\left(u\right)\right)\\ =\frac{d}{du}\left(u^{\frac{1}{2}}\right)\\ =\frac{1}{2}{u}^{\frac{1}{2}-1}\end{array}$

Simplify further and obtain the derivative,

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)=\frac{1}{2}{u}^{\frac{1-2}{2}}\\ =\frac{1}{2}{u}^{\frac{-1}{2}}\\ =\frac{1}{2u^{\frac{1}{2}}}\\ =\frac{1}{2\sqrt{u}}\end{array}$

Substitute $u=f\left(x\right)$ in the above equation,

$g^{\prime }\left(h\left(x\right)\right)=\frac{1}{2\sqrt{f\left(x\right)}}$

Thus, the derivative $g^{\prime }\left(h\left(x\right)\right)$ is $g^{\prime }\left(h\left(x\right)\right)=\frac{1}{2\sqrt{f\left(x\right)}}$.

The derivative of $h\left(x\right)$ is computed as follows,

$\begin{array}{c}{h}^{\prime }\left(x\right)=\frac{d}{dx}\left(f\left(x\right)\right)\\ =f^{\prime }\left(x\right)\end{array}$

Thus, the derivative of $h\left(x\right)$ is $h^{\prime }\left(x\right)=f^{\prime }\left(x\right)$.

Substitute $\frac{1}{2\sqrt{f\left(x\right)}}$ for $g^{\prime }\left(h\left(x\right)\right)$ and $f^{\prime }\left(x\right)$ for $h^{\prime }\left(x\right)$ in equation (2),

$\begin{array}{c}\frac{d}{dx}\left[\sqrt{f\left(x\right)}\right]=\frac{1}{2\sqrt{f\left(x\right)}}\left(f^{\prime }\left(x\right)\right)\\ =\frac{f^{\prime }\left(x\right)}{2\sqrt{f\left(x\right)}}\end{array}$

Thus, the left hand side is equal to the right hand side.

Therefore, the given statement is true.