MATH 241, LECTURE 18
1. Implicit differentiation
Here are the basic steps to use implicit differentiation to find the derivative of ywith respect to xif the
two are related by an equation but not a function.
•Differentiate the equation, treating yas a function of xeven though it is not expressed as one.
Remember to use the chain rule to differentiate y, since we are treating it as a function of x!
•The resulting expression has dy
dx in it, since we have used the chain rule on y. Now do algebra to
solve for it.
•Now that we know a formula for dy
dx , we can plug in values for xand yto find a numerical value
for the derivative.
Example 1. Find dy
dx when exy+4x−3= 5x2. (Do this twice, the second time by taking the natural logarithm
of both sides).
Example 2. Two manufacturers of widgets are in direct competition. Because of the many variables in
pricing and publicity, the number of widgets they sell does not add up to a constant, but 6x2+xy + 5y2=
120,000, where xis the number of widgets sold per day by the first company and yby the second. If both
companies are currently selling 100 widgets, what would be the effect on the first company if the second is
increase sales by ten widgets per day.
Example 3. The Cobb-Douglas model of production uses functions of the form P=xay1−ato model
production level P, where xrepresents an amount of capital and yrepresents an amount labor. Why does
this make sense? Suppose widget production level Pis modeled by x1
4y3
4, where Pis measured in thousands
of units, xin millions of dollars, and yin hundreds of workers. If currently xis 16 and yis 81, what would
dP
dt be if there is currently a move to capitalize further at a rate of $2 million per quarter and increase
workers at a rate of 300 per quarter. What is the expected increase in production per quarter?
Example 4. When the price of gadgets is pdollars each, the manufacturer is willing to supply xhundred
units where x2−2x
5+p−p2= 20. How fast is the supply changing when the price is $5 and increasing at
the rate of 20 cents per week?
1.1. Understanding the derivative of the logarithm function. Now that we have implicit differen-
tiation, we can use the fact that d
dx ex=exto show that
Theorem 5. The derivative of ln(x)with respect to xis 1
x.
Proof: Start with the fact that eln x=x, and differentiate both sides. Remember that we don’t really
know yet what the derivative of ln xis. To take the derivative of the left side, we must use the chain rule;
the derivative of xis straightforward:
d
dx eln x=eln xd
dx ln x=xd
dx ln x,
while d
dx x= 1. So we have xd
dx ln x= 1, so d
dx ln x=1
x.
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