## Section4.2Derivative 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.

### Subsection4.2.1Derivatives of scalar products

We start by differentiating a constant times a function.

#### Example4.2.2.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. $$\displaystyle h'(x)=[\ln(x^7)]'=[7 \ln(x)]'=7[\ln(x)]'=7/x.$$

### Subsection4.2.2Derivatives of sums and differences

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

#### Example4.2.4.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{align*} h'(x)\amp =5[x^{1/2}]'+2[x^{-1/2}]'-7[x^{-3}]'\\ \amp =5/2 x^{-1/2}-x^{-3/2}+21 x^{-4}\\ \amp =\frac{5}{2\sqrt{x}}-\frac{1}{\sqrt{x^3}}+\frac{21}{x^4}\text{.} \end{align*}

#### 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.

\begin{equation*} c*(f' (x_0 )(x-x_0 )+f(x_0 ))=(c f' (x_0 ))(x-x_0 )+(c f(x_0 ))\text{.} \end{equation*}

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{align*} (f' (x_0 )(x-x_0 )+f(x_0 ))\amp \pm(g' (x_0 )(x-x_0 )+g(x_0 ))\\ \amp =( f' (x_0 )\pm g' (x_0 ))(x-x_0 )+(f(x_0 )±g(x_0 )) \text{.} \end{align*}

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.

### Subsection4.2.3Derivatives of products

We now turn our attention to the product of two functions.

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.

#### Example4.2.6.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{align*} (f(x)*g(x))' \amp = f'(x)*g(x)+f(x)*g'(x)\\ \amp = (1)*(x^2 )+(x)*(2x)= 3x^2\text{.} \end{align*}

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

#### Example4.2.7.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.

\begin{equation*} (a x+c)(b x+d)=a b x^2+(a d+b c)x+c d\text{.} \end{equation*}

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

\begin{equation*} (f'(x_0)(x-x_0 )+f(x_0 ))*(g' (x_0 )(x-x_0)+g(x_0 ))\text{,} \end{equation*}

the coefficient of the linear term is

\begin{equation*} f'(x_0 )g(x_0 )+f(x_0)g'(x_0)\text{.} \end{equation*}

### Subsection4.2.4Derivatives of quotients

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

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

#### Example4.2.9.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{align*} (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\text{.} \end{align*}

Both methods give the same answer. Note that the quotient of the derivatives is $$2x\text{,}$$ which is not the same as the derivative of the quotient.

#### Example4.2.10.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. $$\displaystyle f'(x)=\left(\frac{6x^2+100}{x^3+x}\right)'=\frac{(12x)(x^3+x)-(6x^2+100)(3x^2+1)}{(x^3+x)^2}$$

2. $$\displaystyle g'(x)=\left(\frac{1.07^x}{3x+5}\right)'=\frac{(\ln(1.07)1.07^x)(3x+5)-(1.07^x)(3)}{(3x+5)^2}$$

#### Theory and justification.

Following the general method we look at the linear term of the quotient of the linear approximations. However, we need to do an algebraic trick before we can find the linear term. Consider the quotient of two linear expressions:

\begin{equation*} \frac{a+c x}{b+d x}=\frac{(a+c x)(b-d x)}{(b+d x)(b-d x)}=\frac{a b+(c b-a d)x-c d x^2}{b^2 -d^2 x^2}\text{.} \end{equation*}

When $$x$$ is small enough, we get a good approximation by ignoring the $$x^2$$ term. In that approximation, the coefficient of the linear term is $$\frac{(c b-a d)}{b^2}\text{.}$$ Thus, when we take the quotient,

\begin{equation*} \frac{f' (x_0 )(x-x_0 )+f(x_0 )}{g' (x_0 )(x-x_0 )+g(x_0 )}\text{,} \end{equation*}

the coefficient of the linear term is

\begin{equation*} \frac{f'(x_0 )g(x_0 )-f(x_0)g'(x_0 )}{(g(x_0 ))^2}\text{.} \end{equation*}

### Exercises4.2.5Exercises: Derivative Rules for Combinations of Functions Problems

#### Exercise Group.

Use the rules from the last two sections to find the derivatives of the following functions.

##### 1.

$$f(x)=3x^5+7x^4+5x+9\text{.}$$

$$f'(x)=15x^4+28x^3+5.$$

##### 2.

$$g(x)=10x^{123}+19x^{50}-5x^4-2x^{-7}\text{.}$$

##### 3.

$$h(x)=2x+3\sqrt{x}-5\sqrt[3]{x}\text{.}$$

Solution.

Rewrite using exponential notation: $$h(x)=2x+3x^{1/2}-5x^{1/3}$$ Then take the derivative:

\begin{equation*} h'(x)=2+\frac{3}{2} x^{-1/2}-\frac{5}{3} x^{-2/3}\text{.} \end{equation*}
##### 4.

$$k(x)=e^x+8^x+\pi^x-1.07^x\text{.}$$

##### 5.

$$m(x)=(x^3+5x^2-7)(x^2-3)\text{.}$$

Solution.

Product Rule:

\begin{align*} m'(x)\amp =(x^3+5x^2-7)'(x^2-3)+(x^3+5x^2-7) (x^2-3)'\\ \amp=(3x^2+10x)(x^2-3)+(x^3+5x^2-7)(2x)\text{.} \end{align*}
##### 6.

$$n(x)=(10x-9)(x^5+\sqrt{x})\text{.}$$

##### 7.

$$f(x)=e^x \ln(x)\text{.}$$

Solution.

Product Rule:

\begin{equation*} f'(x)=(e^x)' \ln(x)+e^x (\ln(x))'= e^x\ln(x)+e^x \frac{1}{x}=e^x [\ln(x)+\frac{1}{x}]\text{.} \end{equation*}

Note: the equation $$f'(x)= e^x \ln(x)+\frac{e^x}{x}$$ is also acceptable as a solution.

Factoring is sometimes done, but depends on the problem we are trying to solve and what we are trying to do with the derivative.

##### 8.

$$g(x)=1.07^x (10x-15)\text{.}$$

##### 9.

$$h(x)=(a x+b)(c x+d)(e x+f)\text{.}$$

Solution.

If we have a product of 3 functions we take the derivative one at a time:

\begin{align*} h'(x)\amp =(a x+b)' (c x+d)(e x+f)+(a x+b) (c x+d)' (e x+f)+(a x+b)(c x+d) (e x+f)'\\ \amp =a(c x+d)(e x+f)+(a x+b)c (e x+f)+(a x+b)(c x+d)e\text{.} \end{align*}
##### 10.

$$k(x)=1.04^x e^x\text{.}$$

##### 11.

$$m(x)=\frac{(x^2-5x)}{(x^2+3)}\text{.}$$

Solution.

Quotient Rule:

\begin{align*} m'(x)\amp =\frac{(x^2-5x)' (x^2+3)-(x^2-5x) (x^2+3)'}{(x^2+3)^2}\\ \amp =\frac{(2x+5)(x^2+3)-(x^2-5x)(2x)}{(x^2+3)^2}\text{.} \end{align*}

We can simplify this :

\begin{equation*} m'(x)=\frac{2x^3+5x^2+6x+15-(2x^3-10x^2 )}{(x^2+3)^2} =\frac{-5x^2+6x+15}{(x^2+3)^2}\text{.} \end{equation*}
##### 12.

$$n(x)=\frac{(x^2-5x+3)}{e^x}\text{.}$$

##### 13.

$$f(x)=\frac{ln(x))}{(2x+7)}\text{.}$$

Solution.

Quotient Rule:

\begin{align*} f'(x)\amp =\frac{(\ln(x))'(2x+7)-(\ln(x))(2x+7)'}{(2x+7)^2}\\ \amp =\frac{(1/x)(2x+7)-(\ln(x))(2)}{(2x+7)^2}\\ \amp =\frac{x}{x}\frac{ (1/x)(2x+7)-(\ln(x))(2)}{(2x+7)^2}\\ \amp = \frac{2x+7-2x \ln(x)}{x (2x+7)^2}\text{.} \end{align*}
##### 14.

$$g(x)=\frac{(x^3+2x+9)}{1.07^x}\text{.}$$

##### 15.

$$h(x)=\frac{(a x+b)}{(c x+d)}\text{.}$$

Solution.

Quotient Rule:

\begin{align*} h'(x)\amp =\frac{(a x+b)' (c x+d)-(a x+b) (c x+d)'}{(c x+d)^2}\\ \amp = \frac{a(c x+d)-(a x+b)c}{(c x+d)^2} =\frac{acx+ad-(ac x+bc)}{(c x+d)^2}\\ \amp =\frac{ad-bc}{(c x+d)^2}\text{.} \end{align*}
##### 16.

$$k(x)=\frac{((3x+5)(x^2+7))}{(x^2+2x)}\text{.}$$

##### 17.

$$m(x)=\frac{\ln(x)1.04^x}{3x^2+7}\text{.}$$

Solution.

Quotient Rule with embedded Product Rule:

\begin{align*} m'(x)\amp =\frac{(\ln(x)(1.04^x ))'(3x^2+7) -(\ln(x) (1.04^x))(3x^2+7)'}{(3x^2+7)^2}\\ \amp =\frac{(\ln(x)'(1.04^x )+\ln(x)(1.04^x )')(3x^2+7) -(\ln(x) (1.04^x))(3x^2+7)'}{(3x^2+7)^2} \\ \amp =\frac{(\frac{1}{x}(1.04^x )+\ln(x)\ln(1.04)(1.04^x ))(3x^2+7) -(\ln(x) (1.04^x))(6x)}{(3x^2+7)^2} \end{align*}
##### 18.

$$m(x)=\frac{1.06^x e^x}{1.08^x}$$

#### Exercise Group.

For the following problems, use the following data to find the indicated derivative.

 x 0 1 2 3 4 5 6 7 8 9 f(x) 3 5 7 1 9 8 4 2 0 6 f'(x) 7 6 5 4 3 2 1 0 9 8 g(x) 8 4 0 6 2 9 5 1 7 3 g'(x) 6 8 4 2 0 7 9 3 5 1
##### 19.

$$h'(2)\text{,}$$ where $$h(x)=f(x)*g(x)\text{.}$$

Solution.
\begin{align*} h' (x)\amp =f' (x)*g(x)+f(x)*g' (x)\\ h' (2)\amp =f' (2)*g(2)+f(2)*g' (2)\\ \amp =5*0+7*4=28\text{.} \end{align*}
##### 20.

$$h'(5)\text{,}$$ where $$h(x)=f(x)-g(x)\text{.}$$

##### 21.

$$h'(7)\text{,}$$ where $$h(x)=f(x)/g(x)\text{.}$$

Solution.
\begin{align*} h'(x)\amp =\frac{f' (x)g(x)-f(x) g' (x)}{(g(x))^2} \\ h' (7)\amp =\frac{f' (7)g(7)-f(7) g' (7)}{(g(7))^2}\\ \amp =\frac{0*1-2*3}{(1)^2} =-6\text{.} \end{align*}
##### 22.

$$h'(4)\text{,}$$ where $$h(x)=g(x)/f(x)\text{.}$$

#### 23.

The profit function at the widget factory is $$\profit(q)=-q^2+300q-2500\text{.}$$ Find the maximum profit.

Solution.

$$\profit (q)$$ is a downward opening parabola, so the maximum occurs where the derivative (slope of the tangent line) is 0.

Find the critical point: $$\profit' (q)=-2q+300=0\text{,}$$ So $$q = 150\text{.}$$

The maximum profit is

\begin{align*} \profit(150)\amp =-150^2+300(150)-2500\\ \amp =-22,500+45,000-2500=20,000\text{.} \end{align*}

#### 24.

The demand function for a fad item is $$\demand(t)=100t^2 (0.8)^t\text{,}$$ with $$t$$ measured in months. When is the demand the highest?

#### 25.

The cost function for gizmo production is $$\cost(q)=3000+10q-0.02q^2\text{,}$$ for $$q\leq 200\text{.}$$ Find the equation of the line tangent to the cost function at $$q=200\text{.}$$

Solution.

For the tangent line we need a point and a slope, and once we have those we find the equation of the line.

Point: When $$q = 200\text{,}$$

\begin{align*} \cost(200)\amp =3000+10(200)-.02(200)^2\\ =3000+2000-.02*40,000=4200\text{.} \end{align*}

Slope: $$\frac{dcost}{dq}=10-.04q\text{,}$$ so at $$q = 200$$ we have $$\frac{dcost}{dq}=10-.04(200)= 2\text{.}$$

The line: $$C-C_0=m(q-q_0)\text{.}$$

So we have $$C-4200=2(q-200)\text{,}$$ or in slope intercept form: $$C=2q+3800\text{.}$$

#### 26.

The formula for the current value of a particular revenue stream is $$\val(t)=.95^{-t} (3000+25t)\text{.}$$ Find the equation of the line tangent to the cost function at $$t=10\text{.}$$