C.2 Chain Rule - Basic Practice

Solution 1 (quick, the way most people reason).

Think something like: "The function is some stuff to the eighth-power. So the derivative is eight times that same stuff to the seventh power, times the derivative of that stuff." 𝑑𝑓𝑑𝑥=[𝑑𝑓𝑑(stuff), with the same stuff inside]×𝑑𝑑𝑥(stuff) 𝑓(𝑥)=(stuff)8;stuff=3𝑥2−4𝑥+5Then𝑓(𝑥)=𝑑𝑓𝑑𝑥=8(stuff)7⋅𝑑𝑑𝑥(3𝑥2−4𝑥+5)=8(3𝑥2−4𝑥+5)7⋅(6𝑥−4)⟹ (B)✓

Note: You'd never actually write out "stuff = ...." Instead just hold in your head what that "stuff" is, and proceed to write down the required derivatives.

Solution 2 (more formal).

Let's use the first form of the Chain rule above: [𝑓(𝑔(𝑥))]′=𝑓′(𝑔(𝑥))⋅𝑔′(𝑥)=[derivative of the outer function, evaluated at the inner function] × [derivative of the inner function] We have the outer function 𝑓(𝑢) =𝑢8 and the inner function 𝑢 =𝑔(𝑥) =3𝑥2 −4𝑥 +5.

Then 𝑓′(𝑢) =8𝑢7, and 𝑔′(𝑥) =6𝑥 −4. Hence 𝑓′(𝑥)=8𝑢7⋅(6𝑥−4)=8(3𝑥2−4𝑥+5)7⋅(6𝑥−4)⟹ (B)✓

We'll again solve this two ways. The first is the way most experienced people quickly develop the answer, and that we hope you'll soon be comfortable with. The second is more formal.

Solution 1 (quick, the way most people reason).

Think something like: "The function is some stuff to the power of 3. So the derivative is 3 times that same stuff to the power of 2, times the derivative of that stuff." 𝑑𝑓𝑑𝑥=[𝑑𝑓𝑑(stuff), with the same stuff inside]×𝑑𝑑𝑥(stuff) 𝑓(𝑥)=[stuff]3;stuff=tan⁡𝑥Then𝑓(𝑥)=𝑑𝑓𝑑𝑥=3[stuff]2⋅𝑑𝑑𝑥(tan⁡𝑥)=3[tan⁡𝑥]2⋅sec2⁡𝑥=3tan2⁡𝑥⋅sec2⁡𝑥⟹ (D)✓

Note: You'd never actually write out "stuff = ...." Instead just hold in your head what that "stuff" is, and proceed to write down the required derivatives.

Solution 2 (more formal).

Let's use the first form of the Chain rule above: [𝑓(𝑔(𝑥))]′=𝑓′(𝑔(𝑥))⋅𝑔′(𝑥)=[derivative of the outer function, evaluated at the inner function] × [derivative of the inner function] We have the outer function 𝑓(𝑢) =𝑢3 and the inner function 𝑢 =𝑔(𝑥) =tan⁡𝑥.

Then 𝑓′(𝑢) =3𝑢2, and 𝑔′(𝑥) =sec2⁡𝑥. (Recall that (tan⁡𝑥)′ =sec2⁡𝑥.)
Hence 𝑓′(𝑥)=3𝑢2⋅(sec2⁡𝑥)=3[tan⁡𝑥]2⋅sec2⁡𝑥=3tan2⁡𝑥⋅sec2⁡𝑥⟹ (D)✓

Solution 1 (quick, the way most people reason).

Think something like: "The function is √stuff. So the derivative is 121√that same stuff, times the derivative of that stuff."

Solution 2 (more formal).

Let's use the first form of the Chain rule above: [𝑓(𝑔(𝑥))]′=𝑓′(𝑔(𝑥))⋅𝑔′(𝑥)=[derivative of the outer function, evaluated at the inner function] × [derivative of the inner function] We have the outer function 𝑓(𝑢) =√𝑢 and the inner function 𝑢 =𝑔(𝑥) =sin⁡𝑥.
Then (√𝑢)′ =121√𝑢, and (sin⁡𝑥)′ =cos⁡𝑥. Hence 𝑓′(𝑥)=121√𝑢⋅cos⁡𝑥=121√sin⁡𝑥⋅cos⁡𝑥⟹ (C)✓

Solution 1 (quick, the way most people reason).

Think something like: "The function is 𝑒 to the power of some stuff. So the derivative is 𝑒 to the power of exactly the same stuff, times the derivative of that stuff."

Note: You'd never actually write out "stuff = ...." Instead just hold in your head what that "stuff" is, and proceed to write down the required derivatives.

Solution 2 (more formal).

Let's use the first form of the Chain rule above: [𝑓(𝑔(𝑥))]′=𝑓′(𝑔(𝑥))⋅𝑔′(𝑥)=[derivative of the outer function, evaluated at the inner function] × [derivative of the inner function] We have the outer function 𝑓(𝑢) =𝑒𝑢 and the inner function 𝑢 =𝑔(𝑥) =sin⁡𝑥.
Then 𝑓′(𝑢) =𝑒𝑢, and 𝑔′(𝑥) =cos⁡𝑥. Hence 𝑓′(𝑥)=𝑒𝑢⋅cos⁡𝑥=𝑒sin⁡𝑥⋅cos⁡𝑥⟹ (D)✓

Solution 1 (quick, the way most people reason).

Think something like: "The function is 𝑒 to the power of some stuff. So the derivative is 𝑒 to the power of exactly the same stuff, times the derivative of that stuff."

𝑑𝑓𝑑𝑥=[𝑑𝑓𝑑(stuff), with the same stuff inside]×𝑑𝑑𝑥(stuff)

𝑓(𝑥)=𝑒stuff;stuff=𝑥2Then𝑓(𝑥)=𝑑𝑓𝑑𝑥=(𝑒stuff)(𝑑𝑑𝑥(stuff))=𝑒𝑥2⋅2𝑥⟹ (B)✓

Solution 2 (more formal).

Let's use the first form of the Chain rule above: [𝑓(𝑔(𝑥))]′=𝑓′(𝑔(𝑥))⋅𝑔′(𝑥)=[derivative of the outer function, evaluated at the inner function] × [derivative of the inner function] We have the outer function 𝑓(𝑢) =𝑒𝑢 and the inner function 𝑢 =𝑔(𝑥) =𝑥2.

Then 𝑓′(𝑢) =𝑒𝑢, and 𝑔′(𝑥) =2𝑥.

Hence 𝑓′(𝑥)=𝑒𝑢⋅2𝑥=𝑒𝑥2⋅2𝑥⟹ (B)✓