Answer
a) The derivative of $f\left( x \right)=\sqrt{x}$ at x is $f'\left( x \right)=\frac{1}{2\sqrt{x}}$.
b) The slope of the tangent line to the graph of $f\left( x \right)=\sqrt{x}$ at $x=1$ is $f'\left( 1 \right)=\frac{1}{2}$ and at $x=4$ is $f'\left( 4 \right)=\frac{1}{4}$.
Work Step by Step
(a)
Consider the function, $f\left( x \right)=\sqrt{x}$
Now, compute the derivate of $f\left( x \right)=\sqrt{x}$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$
To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)=\sqrt{x}$.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{x+h}-\sqrt{x}}{h}$
Now, multiply and divide with the conjugate of $\sqrt{x+h}-\sqrt{x}$ that is $\sqrt{x+h}+\sqrt{x}$ and simplify using the property $\left( A-B \right)\left( A+B \right)={{A}^{2}}-{{B}^{2}}$.
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( \sqrt{x+h} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}}{h\left( \sqrt{x+h}+\sqrt{x} \right)} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{x+h-x}{h\left( \sqrt{x+h}+\sqrt{x} \right)} \\
\end{align}$
Combine the like terms in the numerator, then divide numerator and denominator by h.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{x+h}+\sqrt{x}}$
Apply the limits and simplify,
$\begin{align}
& f'\left( x \right)=\frac{1}{\sqrt{x+0}+\sqrt{x}} \\
& =\frac{1}{2\sqrt{x}}
\end{align}$
Thus, the derivative of $f\left( x \right)=\sqrt{x}$ at x is $f'\left( x \right)=\frac{1}{2\sqrt{x}}$.
(b)
Consider the function, $f\left( x \right)=\sqrt{x}$
From part (a), the derivative of $f\left( x \right)=\sqrt{x}$ at x is $f'\left( x \right)=\frac{1}{2\sqrt{x}}$.
To compute the slope of the tangent line at $x=1$, substitute $x=1$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( 1 \right)=\frac{1}{2\sqrt{1}} \\
& =\frac{1}{2}
\end{align}$
To compute the slope of the tangent line at $x=4$, substitute $x=4$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( 4 \right)=\frac{1}{2\sqrt{4}} \\
& =\frac{1}{2\cdot 2} \\
& =\frac{1}{4}
\end{align}$
