Precalculus (6th Edition) Blitzer

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 $f'\left( 4 \right)=\frac{1}{4}$.
(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}