Study guides › 0606 › Calculus - Differentiation 1 Maths4U Original · Learn, check, explain
The quotient rule — 0606
Differentiate a quotient and preserve its excluded values.
1. Understand 2. See examples 3. Check answers 4. Explain your working Notes and key methods For \(\displaystyle y= \frac{u}{v}, y'= \frac{u'v- uv'}{v^{2}}\) where \(\displaystyle v\ne 0\). Write the numerator with brackets before simplifying. For a linear-over-linear expression, polynomial division can give a constant plus a reciprocal and an independent derivative check.
The square in the denominator is essential. The original excluded value remains excluded even if another expression simplifies.
Check the conditions Use u′v−uv′ over \(\displaystyle v^{2}\). Keep the original denominator nonzero. Three worked examples Example 1 · The quotient rule for rational functions Let \(\displaystyle f(x)=\frac{5 x + 24}{x + 4}\), with \(x\ne -4\).
Differentiate the given function, showing the rule used. Find the second derivative. Find the equation of the tangent at \(x=0\). Find the height of the horizontal asymptote. Show the reasoning (a) \(\displaystyle \frac{5}{x + 4} - \frac{5 x + 24}{\left(x + 4\right)^{2}}\) follows by differentiating each factor or inner function as required.
Answer: \(\displaystyle f'(x) = \frac{5}{x + 4} - \frac{5 x + 24}{\left(x + 4\right)^{2}}\)
(b) Differentiate the complete first derivative and simplify. \(\displaystyle \frac{2 \left(-5 + \frac{5 x + 24}{x + 4}\right)}{\left(x + 4\right)^{2}}\)
Answer: \(\displaystyle \frac{2 \left(-5 + \frac{5 x + 24}{x + 4}\right)}{\left(x + 4\right)^{2}}\)
(c) The point has height \(6\) and gradient \(\displaystyle - \frac{1}{4}\). Use \(y-y_0=m(x-x_0)\) to obtain \(\displaystyle y=6 - \frac{x}{4}\).
Answer: \(\displaystyle y=6 - \frac{x}{4}\)
(d) Compare the leading coefficients.
\(5\)
Answer: \(5\).
Example 2 · The quotient rule for rational functions Let \(\displaystyle f(x)=\frac{2 x + 15}{x + 5}\), with \(x\ne -5\).
Differentiate the given function, showing the rule used. Find the second derivative. Find the equation of the tangent at \(x=0\). Find the height of the horizontal asymptote. Show the reasoning (a) \(\displaystyle \frac{2}{x + 5} - \frac{2 x + 15}{\left(x + 5\right)^{2}}\) follows by differentiating each factor or inner function as required.
Answer: \(\displaystyle f'(x) = \frac{2}{x + 5} - \frac{2 x + 15}{\left(x + 5\right)^{2}}\)
(b) Differentiate the complete first derivative and simplify. \(\displaystyle \frac{2 \left(-2 + \frac{2 x + 15}{x + 5}\right)}{\left(x + 5\right)^{2}}\)
Answer: \(\displaystyle \frac{2 \left(-2 + \frac{2 x + 15}{x + 5}\right)}{\left(x + 5\right)^{2}}\)
(c) The point has height \(3\) and gradient \(\displaystyle - \frac{1}{5}\). Use \(y-y_0=m(x-x_0)\) to obtain \(\displaystyle y=3 - \frac{x}{5}\).
Answer: \(\displaystyle y=3 - \frac{x}{5}\)
(d) Compare the leading coefficients.
\(2\)
Answer: \(2\).
Example 3 · The quotient rule for rational functions Let \(\displaystyle f(x)=\frac{3 x + 9}{x + 1}\), with \(x\ne -1\).
Differentiate the given function, showing the rule used. Find the second derivative. Find the equation of the tangent at \(x=0\). Find the height of the horizontal asymptote. Show the reasoning (a) \(\displaystyle \frac{3}{x + 1} - \frac{3 x + 9}{\left(x + 1\right)^{2}}\) follows by differentiating each factor or inner function as required.
Answer: \(\displaystyle f'(x) = \frac{3}{x + 1} - \frac{3 x + 9}{\left(x + 1\right)^{2}}\)
(b) Differentiate the complete first derivative and simplify. \(\displaystyle \frac{6 \left(-1 + \frac{x + 3}{x + 1}\right)}{\left(x + 1\right)^{2}}\)
Answer: \(\displaystyle \frac{6 \left(-1 + \frac{x + 3}{x + 1}\right)}{\left(x + 1\right)^{2}}\)
(c) The point has height \(9\) and gradient \(-6\). Use \(y-y_0=m(x-x_0)\) to obtain \(y=9 - 6 x\).
Answer: \(y=9 - 6 x\)
(d) Compare the leading coefficients.
\(3\)
Answer: \(3\).
Original exam-style problems Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.
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