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 5 rules.
Elementary Formulas.
If \(f(x)=a\text{,}\) then \(f'(x)=0\text{.}\)
If \(f(x)=ax\text{,}\) then \(f'(x)=a\text{.}\)
If \(f(x)=a*x^n\text{,}\) then \(f'(x)=a*n*x^{n-1}\text{,}\) for any nonzero number n.
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 \(f'(x)=1/x\)
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.
Claim4.2.1.Scalar multiple rule.
The derivative of \(c*f(x)\) is \(c*f'(x)\text{.}\) In other words,
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.
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.
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.
Claim4.2.5.Product rule.
The derivative of \(f(x)* g(x)\) is \(f'(x)g(x)+f(x)g'(x)\text{.}\) In other words,
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:
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.
For our first example we look at a case that we dane do without the quotient rule by simplifying first. This lets us check our answer. Let \(f(x)=x^2\) and \(g(x)=x\text{.}\) Find the derivative of \(f(x)/g(x)\text{.}\)
Solution.
We start by simplifying. Note that \(f(x)/g(x)=x\text{.}\) Using our rule for monomials, \((f(x)*g(x))'=(x )'=1\text{.}\)
Now we use the quotient rule directly. Using the same rule we see \(f'(x)=2x\text{,}\) and \(g'(x)=1\text{.}\) Using the quotient rule:
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:
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,
1.Reading check, Derivative Rules for Combinations of Functions.
This question checks your reading comprehension of the material is section 4.2, Derivative Rules for Combinations of Functions, of Business Calculus with Excel. Based on your reading, select all statements that are correct. There may be more than one correct answer. The statements may appear in what seems to be a random order.
The derivative of \(f(x)/g(x)\) is \(f'(x)/g'(x)\text{.}\)
The derivative of \(f(x)+g(x)\) is \(f'(x)+g'(x)\text{.}\)
The derivative of \(f(x)*g(x)\) is \(f'(x)*g'(x)\text{.}\)
The derivative of \(f(x)*g(x)\) is \(f'(x)*g(x)+f(x)*g'(x)\text{.}\)
The derivative of \(f(x)/g(x)\) is \(\frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x)^2}\text{.}\)
The derivative of \(f(x)-g(x)\) is \(f'(x)-g'(x)\text{.}\)
The derivative of \(f(x)/g(x)\) is \(\frac{f(x)*g'(x)-f'(x)*g(x)}{(g(x)^2}\text{.}\)
The derivative of \(c*f(x)\) is \(c*f'(x)\text{.}\)
None of the above
Exercises4.2.6Exercises: 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.
\(f(x)=3x^5+7x^4+5x+9\text{.}\)
Answer.
\(f'(x)=15x^4+28x^3+5.\)
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:
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.
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 \(t=10\text{.}\)