4.2: Derivative Rules for Combinations of Functions

In the last section we learned rules to symbolically differentiate some elementary functions. To summarize, we have established 4 rules.

Elementary Formulas:

If f\((x)=x^n\text{,}\) then \(f'(x)=n*x^{(n-1)}\text{,}\) for any real number \(n\text{.}\)

If \(f(x)=e^x\text{,}\) then \(f'(x)=e^x\text{.}\)

If \(f(x)=a^x\text{,}\) then \(f'(x)=a^x \ln(a)\text{.}\)

If \(f(x)=\ln(x)\text{,}\) then \(f'(x)= 1/x\text{.}\)

However we do not yet have a rule for taking the derivative of a function as simple as \(f(x)=x+2\text{.}\) Rather than producing rules for each kind of function, we wish to discover how to differentiate functions obtained by arithmetic on functions we already know how to differentiate. This would let us differentiate functions like \(f(x)=5 x^3+3x^2+1\text{,}\) or \(g(x)=(x+2) 1.03^x\text{,}\) or \(F(x)= \ln(x)/(x+3)\text{,}\) which are built up from our elementary functions. We want rules for multiplying a known function by a constant, for adding or subtracting two known functions, and for multiplying or dividing two known functions.

4.2.1Derivatives of scalar products

We start by differentiating a constant times a function.

Claim \(4.2.1\). Scalar Multiple Rule.

The derivative of \(c*f(x)\) is \(c*f'(x)\text{.}\) In other words,

\[ [c*f(x)]'=c*f'(x)\text{.} \nonumber \]

Example 4.2.2: Derivatives of Constants times Standard Functions.

Find the derivatives of the following functions:

1. \(\displaystyle f(x)=2e^x\)
2. \(\displaystyle g(x)=500(1.05)^x\)
3. \(\displaystyle h(x)=\ln(x^7)\)

Solution

Using our rule

1. \(\displaystyle f'(x)=2[e^x ]'=2e^x.\)
2. \(\displaystyle g'(x)=500[(1.05)^x ]'=500(1.05)^x \ln(1.05).\)
3. \(\displaystyle h'(x)=[\ln(x^7)]'=[7 \ln(x)]'=7[\ln(x)]'=7/x.\)

Derivatives of sums and differences

Next we want to look at the sum or difference of two functions.

Claim \(4.2.3\). Sum and Difference Rule.

The derivative of \(f(x)\pm g(x)\) is \(f'(x)\pm g'(x)\text{.}\) In other words,

\[ [f(x)\pm g(x)]'=f'(x)\pm g'(x)\text{.} \nonumber \]

Example 4.2.4: Derivatives of Sums and Differences of Standard Functions.

Find the derivatives of the following functions:

1. \(\displaystyle f(x)=5x^3+3x^2-7\)
2. \(\displaystyle g(x)=100e^x-1000(1.03)^x\)
3. \(\displaystyle h(x)=5\sqrt{x}+2/\sqrt{x}-7x^{-3}\)

Solution

Using our rule

1. \(\displaystyle f'(x)=[5x^3 ]'+[3x^2 ]'-[7]'=15x^2+6x-0.\)
2. \(\displaystyle g'(x)=100[e^x]'-1000(1.03)^x]'=100e^x-1000(1.03)^x \ln(1.03).\)
3. \begin{gather*} h'(x)=5[x^{1/2}]'+2[x^{-1/2}]'-7[x^{-3}]'\\ =5/2 x^{-1/2}-x^{-3/2}+21 x^{-4}\\ =\frac{5}{2\sqrt{x}}-\frac{1}{\sqrt{x^3}}+\frac{21}{x^4}. \end{gather*}

Theory and justification

The basic argument for all of our rules starts with local linearity. Recall that if \(f(x)\) is differentiable at \(x_0\text{,}\) then in a region around \(x_0\text{,}\) we can approximate \(f(x)\) by a linear function, \(f(x)\approx f'(x_0 )(x-x_0 )+f(x_0)\text{.}\) To find the derivative of a scalar product, sum, difference, product, or quotient of known functions, we perform the appropriate actions on the linear approximations of those functions. We then take the coefficient of the linear term of the result.

For our first rule we are differentiating a constant times a function. Following the general method we look at how we multiply a constant times the linear approximation.

\[ c*(f' (x_0 )(x-x_0 )+f(x_0 ))=(c f' (x_0 ))(x-x_0 )+(c f(x_0 )) \nonumber \]

Taking the coefficient of the linear term gives the scalar multiple rule, the derivative of a constant times a functions is the constant times the derivative of the function.

Next we want to look at the sum or difference of two functions. Following the general method we look at the sum or difference of the linear approximations.

\begin{gather*} (f' (x_0 )(x-x_0 )+f(x_0 ))\pm(g' (x_0 )(x-x_0 )+g(x_0 ))\\ =( f' (x_0 )\pm g' (x_0 ))(x-x_0 )+(f(x_0 )±g(x_0 )) \end{gather*}

Taking the coefficient of the linear term gives the sum or difference rule, the derivative of a sum or difference of two functions is the sum or difference of the derivatives of the functions.

4.2.2Derivatives of products

We turn our attention to the product of two functions.

Claim \(4.2.5\). Product Rule.

The derivative of \(f(x)* g(x)\) is \(f'(x)g(x)+f(x)g'(x)\) In other words,

\[ [f(x)*g(x)]'=f'(x)*g(x)+f(x)*g'(x)\text{.} \nonumber \]

Warning: Note that the derivative of a product is not the product of the derivatives!

We start with an example that we can do by multiplying before taking the derivative. This gives us a way to check that we have the rule correct.

Example 4.2.6: Simple Derivative of a Product.

Let \(f(x)=x\) and \(g(x)=x^2\text{.}\) Find the derivative of \(f(x)*g(x)\text{.}\)

Solution

Note that \(f(x)*g(x)=x^3\text{.}\) Using our rule for monomials \((f(x)*g(x))'=(x^3 )'=3x^2\text{.}\) Using the same rule we see \(f'(x)=1\text{,}\) and \(g'(x)=2x\text{.}\) We can now evaluate using the product rule:

\[ \begin{aligned} (f(x)*g(x))' \amp = f'(x)*g(x)+f(x)*g'(x) \\ \amp = (1)*(x^2 )+(x)*(2x)= 3x^2. \end{aligned} \nonumber \]

Both methods give the same answer. Note that the product of the derivatives is \(2x\) which is NOT the derivative of the product.

Example 4.2.7: General Derivatives of Products.

Find the derivatives of the following functions:

1. \(\displaystyle f(x)=(6x+100)*(1.06)^x.\)
2. \(\displaystyle g(x)=\sqrt{x} \ln(x).\)

Solution

1. \(\displaystyle f'(x)=(6)*(1.06)^x+(6x+100)*(1.06)^x \ln(1.06)\)
2. \(\displaystyle g'(x)=[\sqrt{x}]'\ln(x)+\sqrt{x}[\ln(x)]'= [1/(2\sqrt{x})]\ln(x)+\sqrt{x}[1/x].\)

Theory and justification

Following the general rule we look at the linear term of the product of the linear approximations. Consider the product of two linear expressions.

\[ (a x+c)(b x+d)=a b x^2+(a d+b c)x+c d. \nonumber \]

The coefficient of the linear term is \((a d+b c)\text{.}\) Thus, when we take the product

\[ (f'(x_0)(x-x_0 )+f(x_0 ))*(g' (x_0 )(x-x_0)+g(x_0 )), \nonumber \]

the coefficient of the linear term is

\[ f'(x_0 )g(x_0 )+f(x_0)g'(x_0). \nonumber \]

4.2.3Derivatives of quotients

Finally, we turn our attention to the quotient of two functions.

Claim \(4.2.8\). Quotient Rule.

The derivative of \(\frac{f(x)}{g(x)}\) is \(\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}\) In other words,

\[ \left[\frac{f(x)}{g(x)}\right]'=\frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}\text{.} \nonumber \]

Warning: Once again, note that the derivative of a quotient is NOT the quotient of the derivatives!

Example 4.2.9: Simple Derivative of a Quotient.

Let \(f(x)=x^2\) and \(g(x)=x\text{.}\) Find the derivative of \(f(x)/g(x)\text{.}\)

Solution

Note that \(f(x)/g(x)=x\text{.}\) Using our rule for monomials \((f(x)*g(x))'=(x )'=1\text{.}\) Using the same rule we see \(f'(x)=2x\text{,}\) and \(g'(x)=1\text{.}\) We can now evaluate using the quotient rule:

\[ \begin{aligned} (f(x)/g(x))' \amp = \frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2} \\ \amp = \frac{(2x)*(x )-(x^2)*(1)}{x^2}=\frac{x^2}{x^2}=1. \end{aligned} \nonumber \]

Both methods give the same answer. Note that the quotient of the derivatives is \(2x\text{,}\)

Example 4.2.10: General Derivatives of Quotients.

Find the derivatives of the following functions:

1. \(\displaystyle f(x)=((6x^2+100))/(x^3+x).\)
2. \(\displaystyle g(x)=(1.07)^x/(3x+5).\)

Solution

1. \[ f'(x)=\left(\frac{6x^2+100}{x^3+3}\right)' =\frac{(12x)(x^3+3)-(6x^2+100)(3x^2+1)}{(x^3+2)^2} \nonumber \]