# How do you determine the instantaneous rate of change of y(x) = sqrt(3x + 1) for x = 1?

Mar 25, 2015

Apply the chain rule for derivatives and substitute $x = 1$ in the result.

$\frac{d \left(y \left(x\right)\right)}{\mathrm{dx}} = \left(\frac{1}{2}\right) {\left(3 x + 1\right)}^{- \frac{1}{2}} \cdot \left(3\right)$

$\frac{d \left(y \left(1\right)\right)}{\mathrm{dx}} = \frac{3}{2 \sqrt{3 + 1}} = \frac{3}{4}$

$\frac{d \left(y \left(1\right)\right)}{\mathrm{dx}}$ is the instantaneous rate of change of $y \left(x\right)$ at $x = 1$

Mar 25, 2015

If you need to do this from the definition, the details will depend on your textbook and instructor's choice of definition:

You'll have one of:

${\lim}_{x \rightarrow 1} \frac{y \left(x\right) - y \left(1\right)}{x - 1}$ or ${\lim}_{h \rightarrow 0} \frac{y \left(1 + h\right) - y \left(1\right)}{h}$

Definition 1:

${\lim}_{x \rightarrow 1} \frac{y \left(x\right) - y \left(1\right)}{x - 1} = {\lim}_{x \rightarrow 1} \frac{\sqrt{3 x + 1} - \sqrt{3 \left(1\right) + 1}}{x - 1}$

$= {\lim}_{x \rightarrow 1} \frac{\sqrt{3 x + 1} - 2}{x - 1}$

Notice that substitution leads to indeterminate form: $\frac{0}{0}$. The trick (technique) to try in this case is to rationalize the numerator. Multiply numerator and denominator by the conjugate of the numerator. (Multiply the top and bottom by $\left(\sqrt{3 x + 1} + 2\right)$

${\lim}_{x \rightarrow 1} \frac{y \left(x\right) - y \left(1\right)}{x - 1} = {\lim}_{x \rightarrow 1} \left[\frac{\left(\sqrt{3 x + 1} - 2\right)}{x - 1} \frac{\left(\sqrt{3 x + 1} + 2\right)}{\left(\sqrt{3 x + 1} + 2\right)}\right]$

=lim_(xrarr1)(sqrt(3x+1)^2-2^2)/((x-1)(sqrt(3x+1) + 2)

=lim_(xrarr1)(3x+1-4)/((x-1)(sqrt(3x+1) + 2)

=lim_(xrarr1)(3x-3)/((x-1)(sqrt(3x+1) + 2)

=lim_(xrarr1)(3(x-1))/((x-1)(sqrt(3x+1) + 2) (Still form $\frac{0}{0}$)

$= {\lim}_{x \rightarrow 1} \frac{3}{\sqrt{3 x + 1} + 2}$ (No longer form $\frac{0}{0}$)

$= \frac{3}{\sqrt{3 \left(1\right) + 1} + 2} = \frac{3}{2 + 2} = \frac{3}{4}$

Definition 2:

${\lim}_{h \rightarrow 0} \frac{y \left(1 + h\right) - y \left(1\right)}{h} = {\lim}_{h \rightarrow 0} \frac{\sqrt{3 \left(1 + h\right) + 1} - \sqrt{3 \left(1\right) + 1}}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\sqrt{3 h + 4} - 2}{h}$

Notice that substitution leads to indeterminate form: $\frac{0}{0}$. The trick (technique) to try in this case is to rationalize the numerator. Multiply numerator and denominator by the conjugate of the numerator. (Multiply the top and bottom by $\left(\sqrt{3 h + 4} + 2\right)$

${\lim}_{h \rightarrow 0} \frac{y \left(1 + h\right) - y \left(1\right)}{h} = {\lim}_{h \rightarrow 0} \frac{\left(\sqrt{3 h + 4} - 2\right)}{h} \frac{\left(\sqrt{3 h + 4} + 2\right)}{\left(\sqrt{3 h + 4} + 2\right)}$

$= {\lim}_{h \rightarrow 0} \frac{\left({\sqrt{3 h + 4}}^{2} - {2}^{2}\right)}{h \left(\sqrt{3 h + 4} + 2\right)}$

$= {\lim}_{h \rightarrow 0} \frac{3 h + 4 - 4}{h \left(\sqrt{3 h + 4} + 2\right)}$ , (still form $\frac{0}{0}$)

$= {\lim}_{h \rightarrow 0} \frac{3 h}{h \left(\sqrt{h + 4} + 2\right)}$ , (still form $\frac{0}{0}$)

$= {\lim}_{h \rightarrow 0} \frac{3}{\sqrt{h + 4} + 2}$ , (no longer form $\frac{0}{0}$)

$= \frac{3}{\sqrt{0 + 4} + 2} = \frac{3}{2 + 2} = \frac{3}{4}$