C.4 Chain Rule - Used to Find Yet More Derivatives
On this screen, we're going to use the Chain Rule to find yet more derivatives, including 𝑎𝑥,sec𝑥,cot𝑥, and more. This will both (1) give you even more crucial practice at using the Chain Rule, and (2) will let you find the derivative of whatever functions you encounter, plus learn a cool fact about even and odd functions.
Derivative of 𝑎𝑥
Let's see first how we can easily extend our new knowledge that 𝑑𝑑𝑥𝑒𝑢=𝑒𝑢𝑑𝑢𝑑𝑥 to find the derivative of 𝑎𝑥, now much more easily than we could before we had the Chain Rule as a tool.
Practice Problem #1: Derivative of 𝑎𝑥
Show that 𝑑𝑑𝑥𝑎𝑥=𝑎𝑥ln𝑎, where 𝑎 is a constant and 𝑎>0. (For example, 𝑑𝑑𝑥2𝑥=2𝑥ln2.)Hint: Start with 𝑎𝑥=𝑒ln𝑎𝑥, and use 𝑑𝑑𝑥𝑒𝑢=𝑒𝑢⋅𝑑𝑢𝑑𝑥.
Start from the hint, and recall that ln𝑎𝑥=𝑥ln𝑎:𝑑𝑑𝑥𝑎𝑥=𝑑𝑑𝑥(𝑒ln𝑎𝑥)[Recallthatln𝑎𝑥=𝑥ln𝑎]=𝑑𝑑𝑥(𝑒𝑥ln𝑎)[Recall𝑑𝑒𝑢𝑑𝑥=𝑒𝑢𝑑𝑢𝑑𝑥.Here𝑢=𝑥ln𝑎.]=𝑒𝑥ln𝑎⋅(𝑑𝑑𝑥(𝑥ln𝑎))[Rememberthat𝑎isaconstant,soln𝑎isalsoaconstant.]=𝑒𝑥ln𝑎⋅ln𝑎(𝑑𝑑𝑥(𝑥))=𝑒𝑥ln𝑎⋅ln𝑎=𝑒ln𝑎𝑥⋅ln𝑎=𝑎𝑥ln𝑎✓
By the way, notice that this result still applies when 𝑎=𝑒:𝑑𝑑𝑥𝑒𝑥=𝑒𝑥ln(𝑒)1=𝑒𝑥You probably have firmly in mind that 𝑑𝑑𝑥𝑒𝑥=𝑒𝑥, and so just need to remember that if the base of the exponential a is anything other than e, you have to multiply the derivative by ln𝑎. For instance, 𝑑𝑑𝑥3𝑥=3𝑥ln3.
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By the way, notice that the result of Problem #1, 𝑑𝑑𝑥𝑎𝑥=𝑎𝑥ln𝑎, still applies when 𝑎=𝑒:𝑑𝑑𝑥𝑒𝑥=𝑒𝑥ln(𝑒)1=𝑒𝑥
You probably have firmly in mind that 𝑑𝑑𝑥𝑒𝑥=𝑒𝑥, and so just need to remember that if the base of the exponential a is anything other than e, you have to multiply the derivative by ln𝑎. For instance, 𝑑𝑑𝑥3𝑥=3𝑥ln3.
Quotient Rule from Chain and Product Rules
Let's next use the Chain Rule as an alternate way to develop the Quotient Rule, starting from the Product Rule.
Practice Problem #2: Quotient Rule from Chain and Product Rules
Use the Chain Rule and the Product Rule to develop the Quotient Rule. Start from 𝑓(𝑥)𝑔(𝑥)=𝑓(𝑥)(𝑔(𝑥))−1.
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The next problem develops a slick insight into the derivatives of even and odd functions.
Practice Problem #4: Derivatives of even and odd functions
Show the following.
(a)
Prove that the derivative of an even function is an odd function. [Recall that for an even function, 𝑓(−𝑥)=𝑓(𝑥).]
(b)
Prove that the derivative of an odd function is an even function. [Recall that for an odd function, 𝑓(−𝑥)=−𝑓(𝑥).]
(a)If 𝑓(𝑥) is even, then 𝑓′(−𝑥)=−𝑓′(𝑥).
(b)If 𝑓(𝑥) is odd, then 𝑓′(−𝑥)=𝑓′(𝑥).
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Since we're trying to show something about the derivative of an even function, start with its definition and take the derivative:𝑓(−𝑥)=𝑓(𝑥)𝑑𝑑𝑥[𝑓(−𝑥)]=𝑑𝑑𝑥[𝑓(𝑥)]𝑓′(−𝑥)⋅[𝑑𝑑𝑥(−𝑥)]=𝑓′(𝑥)𝑓′(−𝑥)⋅(−1)=𝑓′(𝑥)−𝑓′(−𝑥)=𝑓′(𝑥)𝑓′(−𝑥)=−𝑓′(𝑥)The last line shows, as requested, that the derivative of 𝑓, 𝑓′, is itself an odd function: 𝑓′(−𝑥)=−𝑓′(𝑥). ✓A less formal approach, using "stuff":We can view the function 𝑓(−𝑥) as being 𝑓(stuff), where ``stuff''=−𝑥.Then
𝑓(stuff)=𝑓(𝑥)𝑑𝑑𝑥[𝑓(stuff)]=𝑑𝑑𝑥[𝑓(𝑥)]𝑓′(stuff)⋅[𝑑𝑑𝑥(stuff)]=𝑓′(𝑥)𝑓′(−𝑥)⋅[𝑑𝑑𝑥(−𝑥)]=𝑓′(𝑥)𝑓′(−𝑥)⋅(−1)=𝑓′(𝑥)𝑓′(−𝑥)=−𝑓′(𝑥)
The last line shows, as requested, that the derivative of 𝑓, 𝑓′, is itself an odd function: 𝑓′(−𝑥)=−𝑓′(𝑥). ✓
Since we're trying to show something about the derivative of an odd function, start with its definition and take the derivative:
𝑓(−𝑥)=−𝑓(𝑥)𝑑𝑑𝑥[𝑓(−𝑥)]=𝑑𝑑𝑥[−𝑓(𝑥)]𝑓′(−𝑥)⋅[𝑑𝑑𝑥(−𝑥)]=−𝑓′(𝑥)𝑓′(−𝑥)⋅(−1)=−𝑓′(𝑥)𝑓′(−𝑥)=𝑓′(𝑥)
The last line shows, as requested, that the derivative of 𝑓, 𝑓′, is itself an even function: 𝑓′(−𝑥)=𝑓′(𝑥). ✓ A less formal approach, using "stuff":
We can view the function 𝑓(−𝑥) as being 𝑓(stuff), where ``stuff''=−𝑥. Then
𝑓(stuff)=−𝑓(𝑥)𝑑𝑑𝑥[𝑓(stuff)]=𝑑𝑑𝑥[−𝑓(𝑥)]𝑓′(stuff)⋅[𝑑𝑑𝑥(stuff)]=−𝑓′(𝑥)𝑓′(−𝑥)⋅[𝑑𝑑𝑥(−𝑥)]=−𝑓′(𝑥)𝑓′(−𝑥)⋅(−1)=−𝑓′(𝑥)𝑓′(−𝑥)=𝑓′(𝑥)
The last line shows, as requested, that the derivative of 𝑓, 𝑓′, is itself an even function: 𝑓′(−𝑥)=𝑓′(𝑥). ✓
This screen concludes our focus on the Chain Rule, though of course we'll be using it often as we proceed.
For now, what are your thoughts about using this crucial rule? What advice would you give to someone just starting to learn what it's about and how to use it? What remaining questions do you have? Please post on the Forum and let the Community know!
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