Expert Answer to "How do you determine the instantaneous rate of..."

Raven Gardner

Answered question

2022-04-17

How do you determine the instantaneous rate of change of

y (x) = \sqrt{3 x + 1}

for x=1?

Answer & Explanation

Marin Lowe

Beginner2022-04-18Added 18 answers

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

\frac{d (y (x))}{d x} = (\frac{1}{2}) {(3 x + 1)}^{- \frac{1}{2}} \cdot (3)

\frac{d (y (1))}{d x} = \frac{3}{2 \sqrt{3 + 1}} = \frac{3}{4}

\frac{d (y (1))}{d x}

is the instantaneous rate of change of y(x) at x=1

cloirdxti

Beginner2022-04-19Added 12 answers

lim_{h \to 0} \frac{y (1 + h) - y (1)}{h} = lim_{h \to 0} \frac{\sqrt{3 (1 + h) + 1} - \sqrt{3 (1) + 1}}{h}

= lim_{h \to 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

(\sqrt{3 h + 4} + 2)

lim_{h \to 0} \frac{y (1 + h) - y (1)}{h} = lim_{h \to 0} \frac{(\sqrt{3 h + 4} - 2)}{h} \frac{(\sqrt{3 h + 4} + 2)}{(\sqrt{3 h + 4} + 2)}

= lim_{h \to 0} \frac{({\sqrt{3 h + 4}}^{2} - 2^{2})}{h (\sqrt{3 h + 4} + 2)}

= lim_{h \to 0} \frac{3 h + 4 - 4}{h (\sqrt{3 h + 4} + 2)}

, (still form

\frac{0}{0}

)

= lim_{h \to 0} \frac{3 h}{h (\sqrt{h + 4} + 2)}

, (still form

\frac{0}{0}

)

= lim_{h \to 0} \frac{3 h}{\sqrt{h + 4} + 2}

, (no longer form

\frac{0}{0}

)

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

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