Calculus

Differentiation of Functions

The Product and Quotient Rules

Page 1
Problems 1-3
Page 2
Problems 4-18

The Product Rule

Let \(u\left( x \right)\) and \(v\left( x \right)\) be differentiable functions. Then the product of the functions \(u\left( x \right)v\left( x \right)\) is also differentiable and
\[{\left( {uv} \right)^\prime } = u’v + uv’.\] We prove the above formula using the definition of the derivative. For this we find the increment of the functions \({uv}\) assuming that the argument changes by \(\Delta x\):
\[{\Delta \left( {uv} \right) }={ u\left( {x + \Delta x} \right)v\left( {x + \Delta x} \right) }-{ u\left( x \right)v\left( x \right).}\] Take into account that
\[
{u\left( {x + \Delta x} \right) = u\left( x \right) + \Delta u,\;\;\;}\kern0pt
{v\left( {x + \Delta x} \right) = v\left( x \right) + \Delta v,}
\] where \(\Delta u\) and \(\Delta v\) are the increments, respectively, of the functions \(u\) and \(v\). Omitting for brevity the argument \(x\) of the functions \(u\) and \(v\), we can write the increment \(\Delta \left( {uv} \right)\) in the following form:
\[\require{cancel}
{\Delta \left( {uv} \right) = \left( {u + \Delta u} \right)\left( {v + \Delta v} \right) – uv }
= {\cancel{uv} + u\Delta v + v\Delta u + \Delta u\Delta v – \cancel{uv} }
= {u\Delta v + v\Delta u + \Delta u\Delta v.}
\] We proceed to calculate the derivative of the product using the properties of limits:
\[
{{\left( {uv} \right)^\prime } = {\lim\limits_{\Delta x \to 0} \frac{{\Delta \left( {uv} \right)}}{{\Delta x}}} }
= {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v + v\Delta u + \Delta u\Delta v}}{{\Delta x}} }
= {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v}}{{\Delta x}} + \lim\limits_{\Delta x \to 0} \frac{{v\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v.}
\] In the first limit the function \(u\) does not depend on the increment \(\Delta x\). Therefore, it can be taken outside the limit sign. The same applies to the function \(v\) in the second term. We calculate separately the limit \(\lim\limits_{\Delta x \to 0} \Delta v:\)
\[
{\lim\limits_{\Delta x \to 0} \Delta v }={ \lim\limits_{\Delta x \to 0} \left( {\frac{{\Delta v}}{{\Delta x}} \cdot \Delta x} \right) }
= {\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta x }={ v’ \cdot 0 = 0.}
\] Thus, the derivative of the product is given by
\[
{{\left( {uv} \right)^\prime } = {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v}}{{\Delta x}} + \lim\limits_{\Delta x \to 0} \frac{{v\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v }}
= {u\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}} + v\lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v }
= {uv’ + vu’ + u’ \cdot 0 }
= {u’v + uv’.}
\]

Important: The derivative of the product is NOT equal to the product of the derivatives!

From this formula, it is easy to obtain an expression for the derivative of the function \(kf\left( x \right)\), where \(k\) is a constant:
\[
{{\left( {kf\left( x \right)} \right)^\prime } = {k’f\left( x \right) + kf’\left( x \right) }}
= {0 \cdot f\left( x \right) + kf’\left( x \right) }
= {kf’\left( x \right),}
\] so that a constant factor can be taken out of the sign of derivative.

The Quotient Rule

Let \(u\left( x \right)\) and \(v\left( x \right)\) be again differentiable functions. Then, if \(v\left( x \right) \ne 0\), the derivative of the quotient of these functions is calculated by the formula
\[{\left( {\frac{u}{v}} \right)^\prime } = {\frac{{u’v – uv’}}{{{v^2}}}.}\] This formula is proved similarly. The increment of the quotient can be written as
\[
{\Delta \left( {\frac{u}{v}} \right) }={ \frac{{u + \Delta u}}{{v + \Delta v}} – \frac{u}{v} }
= {\frac{{\left( {u + \Delta u} \right)v – u\left( {v + \Delta v} \right)}}{{v\left( {v + \Delta v} \right)}} }
= {\frac{{\cancel{uv} + v\Delta u – \cancel{uv} – u\Delta v}}{{v\left( {v + \Delta v} \right)}} }
= {\frac{{v\Delta u – u\Delta v}}{{v\left( {v + \Delta v} \right)}}.}
\] The derivative of the quotient is expressed as follows:
\[
{{\left( {\frac{u}{v}} \right)^\prime } }={ \lim\limits_{\Delta x \to 0} \frac{{\Delta \left( {\frac{u}{v}} \right)}}{{\Delta x}} }
= {\lim\limits_{\Delta x \to 0} \frac{{\frac{{v\Delta u – u\Delta v}}{{{v^2} + v\Delta v}}}}{{\Delta x}} }
= {\lim\limits_{\Delta x \to 0} \frac{{v\frac{{\Delta u}}{{\Delta x}} – u\frac{{\Delta v}}{{\Delta x}}}}{{{v^2} + v\Delta v}}.}
\] Further, using the properties of limits, we find:
\[
{{\left( {\frac{u}{v}} \right)^\prime } }={ \lim\limits_{\Delta x \to 0} \frac{{v\frac{{\Delta u}}{{\Delta x}} – u\frac{{\Delta v}}{{\Delta x}}}}{{{v^2} + v\Delta v}} }
= {\frac{{v\lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} – u\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}}}}{{\lim\limits_{\Delta x \to 0} {v^2} + \lim\limits_{\Delta x \to 0} \left( {v\Delta v} \right)}} }
= {\frac{{u’v – uv’}}{{{v^2} + v\lim\limits_{\Delta x \to 0} \Delta v}}.}
\] Taking into account that \({\lim\limits_{\Delta x \to 0} \Delta v} = 0\) we obtain the final expression for the derivative of the quotient of two functions:
\[{\left( {\frac{u}{v}} \right)^\prime } = \frac{{u’v – uv’}}{{{v^2}}}.\]

Solved Problems

Click on problem description to see solution.

 Example 1

Find the derivative of the function \({y = {\large\frac{2}{x}\normalsize}}\).

 Example 2

Find the derivative of a power function with the negative exponent \(y = {x^{ – n}}\).

 Example 3

Calculate the derivative \(y = \tan x\) using the quotient rule.

 Example 4

Let \(y = {\sin ^2}x\). Differentiate this function without using the chain rule.

 Example 5

Find the derivative of the secant function \(y = \sec x\).

 Example 6

Find the derivative of the function \(y = {e^x}\cos x\).

 Example 7

Find the derivative of the function \(y = {\large\frac{{{e^x} – 1}}{{{e^x} + 1}}\normalsize}\) and calculate its value at \(x = 0\).

 Example 8

Find the derivative of the function \(y = {x^2}\sin x\).

 Example 9

Given the function \(z = \left( {{x^2} + 1} \right)\left( {x – 1} \right)\). Find the value of its derivative at \(x = -1\).

 Example 10

Derive a formula for the derivative of the product of three functions.

 Example 11

Find the derivative of the function \(y = {\large\frac{{2x}}{{1 – {x^2}}}\normalsize}\).

 Example 12

Calculate the derivative of the function \(y = {\large\frac{{1 + \cos x}}{{\sin x}}\normalsize}\).

 Example 13

Differentiate the function \(y = \left( {3 – 2x} \right)\left( {2 – 3x} \right)\).

 Example 14

Differentiate the function \(y = {x^3}{2^x}\).

 Example 15

Find the derivative of the linear fractional function \(y = {\large\frac{{ax + b}}{{cx + d}}\normalsize}\).

 Example 16

Calculate the derivative of the function \(y = {\large\frac{{3x – 1}}{{{x^4}}}\normalsize}\).

 Example 17

Calculate the derivative of the following function: \(y = {\large\frac{{\sqrt x – 1}}{{\sqrt x + 1}}\normalsize}\).

 Example 18

Derive a formula for the derivative of the function \(f\left( x \right) = {\large\frac{{u\left( x \right)v\left( x \right)}}{{w\left( x \right)}}\normalsize}\).

Example 1.

Find the derivative of the function \({y = {\large\frac{2}{x}\normalsize}}\).

Solution.

Using the quotient rule, we have
\[
{y’\left( x \right) = {\left( {\frac{2}{x}} \right)^\prime } }
= {\frac{{2′ \cdot x – 2 \cdot x’}}{{{x^2}}} }
= {\frac{{0 \cdot x – 2 \cdot 1}}{{{x^2}}} }
= { – \frac{2}{{{x^2}}}.}
\]

Example 2.

Find the derivative of a power function with the negative exponent \(y = {x^{ – n}}\).

Solution.

We write the function in the form \(y\left( x \right) = {\large\frac{1}{{{x^n}}}\normalsize}\) and use the quotient rule.
\[
{y’\left( x \right) = {\left( {\frac{1}{{{x^n}}}} \right)^\prime } }
= {\frac{{1′ \cdot {x^n} – 1 \cdot {{\left( {{x^n}} \right)}^\prime }}}{{{{\left( {{x^n}} \right)}^2}}} }
= {\frac{{0 \cdot {x^n} – n{x^{n – 1}}}}{{{x^{2n}}}} }
= { – \frac{n}{{{x^{2n – n + 1}}}} }
= { – \frac{n}{{{x^{n + 1}}}}.}
\]

Example 3.

Calculate the derivative \(y = \tan x\) using the quotient rule.

Solution.

We can write the tangent function as \(\tan x = {\large\frac{{\sin x}}{{\cos x}}\normalsize}\). Then
\[
{y’\left( x \right) }={ {\left( {\tan x} \right)^\prime } = {\left( {\frac{{\sin x}}{{\cos x}}} \right)^\prime } }
= {\frac{{{{\left( {\sin x} \right)}^\prime }\cos x – \sin x{{\left( {\cos x} \right)}^\prime }}}{{{{\left( {\cos x} \right)}^2}}}.}
\] As \({\left( {\sin x} \right)^\prime } = \cos x\), \({\left( {\cos x} \right)^\prime } = -\sin x,\) the derivative is given by
\[
{{\left( {\tan x} \right)^\prime } }={ \frac{{\cos x \cdot \cos x – \sin x \cdot \left( { – \sin x} \right)}}{{{{\cos }^2}x}} }
= {\frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} }={ \frac{1}{{{{\cos }^2}x}}.}
\]

Page 1
Problems 1-3
Page 2
Problems 4-18