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## Calculus 1

### Course: Calculus 1>Unit 3

Lesson 8: Differentiation using multiple rules

# Differentiating using multiple rules: strategy

AP.CALC:
FUN‑3 (EU)
How to analyze the structure of an elaborate expression do determine which derivative rules to use, and (not less important) in what order.

## Want to join the conversation?

• In the second expression d/dx [sin(x^2 + 5) cos (x)] , wouldn't sin(x^2+5) be a composite function with sin (x) be the outer and x^2 + 5 being the inner functions?
• Yep, Sal even mentions that you would have to solve that part -- if you were actually trying to, for this video only talks about the strategy -- using the chain rule. So you're absolutely right.
• Is the actual solution to the first problem the following?

cos((x^2+5)(cos (x))[2x cos(x)-(x^2+5)sin(x)

I know solutions are not really the point of this video, but I appreciate all the practice I can get
• That's right! I got that answer and thought "No way this is right lol it's way too complex...", so I went to an online calculator and it gave me the same answer! Congrats!
• If a function is the product of two quotients, where would you start? Would it be enough to just do the product rule?
• What you could do is rewrite the function as a single quotient and then use the quotient rule.
• Conceptually i'm comfortable but I am having a computing issue with a practice question. It's not related to multiple rule differentiation, so someone can remove if it shouldn't belong here.

We are doing product rule on three expressions and after differentiating, wind up with this. 2⋅csc(x)⋅sec(x)+2x−csc(x)cot(x)⋅sec(x)+2x⋅csc(x)⋅sec(x)tan(x)

Fine. No problem. But it ends up simplifying to this:
2.csc(x)sec(x)−2x.csc^2(x)+2x.sec^2(x)​

Where did the cot(x) and tan(x) disappear to? I realize this is a trig question and probably a very stupid one, but it's driving me bananas.
• Using the trig definitions, cot(x) = cos(x)/sin(x) and sec(x) = 1/cos(x). Using this we can simplify the third term, -csc(x)cot(x)*sec(x).

-csc(x)*cot(x)*sec(x)
= -csc(x)*(cos(x)/sin(x))*(1/cos(x))
= -csc(x)*(1/sin(x))
= -csc(x)*csc(x)
= -csc^2(x)

We also know that tan(x) = sin(x)/cos(x) and csc(x) = 1/sin(x). We can use this to simplify the final term, 2x*csc(x)*sec(x)*tan(x).

2x*csc(x)*sec(x)*tan(x)
= 2x
(1/sin(x))*sec(x)*(sin(x)/cos(x))
= 2x*sec(x)*(1/cos(x))
= 2x*sec(x)*sec(x)
= 2x*sec^2(x)

(I apologize for the formatting, but I cannot seem to fix the bold text.)
• d\dx (3x² · √̅5̅x̅+̅3̅ According to Sal, you start from the outside, which would be the product rule. f´(x)g(x) + f(x)g´(x).
= (6x · √̅5̅x̅+̅3̅ )+3x²/2√̅5̅x̅+̅3̅.
Then for the inner part, use the chain rule which is f´(g(x))g´(x).
=(1/2√̅5̅x̅+̅3̅ )(5).
Will somebody please take a minute and tell me if I got it right, so far? I didn't include the 3x² in the chain rule part. Not too sure about it, and the videos up to now don't elaborate on it.
(1 vote)
• That is almost correct. The correct derivative would be 6x√(5x + 3) + 15x²/2√(5x + 3).
The application of the product rule and chain rule were both correct. However, in your final answer, you forgot to multiply by 5, the "g'(x)" in the chain rule. Hope that I helped.