Question

derivative of y = sin(tan2x)

Answer 1

@mathmale

Answer 2

First, a quick review.
The outermost / most important function here is the sine.

Think of the input to this sine functin as u, a separate function of x, which in this case happens to be tan 2x.

You with me so far?

Answer 3

wait

Answer 4

what rule to be apply in this problem?

Answer 5

1) derivative of the sine function
2) derivative of the tangent function
3) two applications of the chain rule

Answer 6

oh its hard to solve :3

Answer 7

Your outermost function is the sine: y = sin ( tan 2x ) ) looks like y = sin u.

Your inner function (the argument of the sine) is u = tan 2x.

Answer 8

what should i do 1st?

Answer 9

We're going to do some review first.

What is the derivative of the sine function? What is the derivative of y = sin x?

Answer 10

y' = cos x

Answer 11

Good.

Suppose y = tan x. What is the derivative, y ' , of this function? If possible look it up before you answer.

Answer 12

y' = sec^2 x ?

Answer 13

Good.
Making progress!

OK. Look at y = sin (tan x ). (I did not forget the '2x')

this looks like y = sin u, where u = tan x, right?

Since u is a separate function of x, we must use the Chain rule. First, we differentiate the "sin u" and then we multiply the result by the derivative of u.

Clear, or need mor explaination or examples?

Answer 14

can u give some example?

Answer 15

Let's start with y = sin ( tan x ).
Use these rules:

1) the derivative of the sine function is the cosine function: [ sin x ] ' = cos x
and [ sin u ] ' = cos u.
2) If u happens to be a function of x (which is the case in the problem you've posted), then we have to apply the CHAIN RULE and differentiate u as well, with respect to x:

[ sin u ] ' = [ cos u ] * u '

Answer 16

Example: y = sin ( cos x ); the derivative is found by applying the Chain Rule:

y ' = cos ( cos x ) * [ cos x ] ' = cos ( cos x ) * ( -sin x ). You can't simplify that further. Leave it as is.

Answer 17

Let's go back to y = sin ( tan x ). Here u = tan x and u ' = ( sec x )^2.

Therefore, y ' = cos ( tan x ) * ( sec x )^2. That's the answer. Questions?

Answer 18

the u' = sec x ^ 2 or u ' = sec^2 x ? which is true?

Answer 19

If u = tan x, then
u ' = (sec x)^2.

Answer 20

okay whats next?

Answer 21

First: decide whether or not you're OK with y ' = cos ( tan x ) * ( sec x )^2. as the derivative of y = sin (tan x). If you are, fine; if not, tell me what needs further explanation.

Answer 22

i understand why it become y' = cos ( tan x ) * ( sec x )^2

Answer 23

OK, great. Now we're almost ready to solve the original problem that you've posted:

derivative of y = sin(tan2x)


What is the derivative of tan 2x ? Hint: Be certain to use the Chain Rule.

Answer 24

what is the equivalent of tan 2x?

Answer 25

There's a formula for that, a trig identity, and I'd be happy to help you find it, but right now you don't really need that trig identity.

Hint: the derivative of tan x is (sec x)^2.

Go one step further and find the derivative of tan 2x using the chain rule.

Answer 26

tan 2x' = 2sec^2 (2x) ?

Answer 27

Very close. WRite [tan 2x] ' instead of tan 2x ' for clarity.

Answer 28

ok ok

Answer 29

The right side of your equation is entirely appropriate; you have enclosed that '2x' inside parentheses as you should. Where did the 2 in front of [sec(2x}] come from? It is correct, but I'd like to know why you write your answer that way.

Answer 30

i'm only substitute the value of 2

Answer 31

Actually, Jay, you are differentiating the argument '2x'; the derivative of that is 2. According to the Chain Rule you must differentiate (2x) and multiply the derivative of tan (2x) by your result (2).

Answer 32

Please fix your tan 2x' = 2sec^2 (2x) Hint: review our discussion, above.

Answer 33

[tan 2x] ' = 2sec^2 (2x)

Answer 34

Great. Perfect. You're done.

Answer 35

whats next

Answer 36

can you summarize the flow of solution ?

Answer 37

Look at what you initially posted: "derivative of y = sin(tan2x)"

Then look at what you ended up with: [tan 2x] ' = 2sec^2 (2x)

Explain this in words.

Answer 38

What was the goal of this math prob lem?

Answer 39

to get the derivative of y

Answer 40

Right, and you have done that. The only thing missing is a label for y our result:\[\frac{ dy }{ dx }=[\tan 2x] ' = 2\sec^2 (2x) \]

Answer 41

or you could have used y ' as your label. In either case you are done with this differentiation and your result is correct.

Answer 42

what could be happen to sin?

Answer 43

Review:

"derivative of y = sin(tan2x)"

1. differentiate sin (tan 2x) with respect to tan 2x: you'll get cos (tan 2x).
2. differentiate that tan 2x; you'll get 2(sec 2x)^2.
3. multiply together the results of (1) and (2), above.

Answer: y ' = cos (tan 2x) * 2(sec 2x)^2.

Answer 44

It's essential that you be able to do this yourself, without help. You can ask all the questions you want. Review our discussion.
List, review and apply the rules of differentiation I've shared with you:

Product rule
Chain rule
Implicit differentiation
Power rule
Power rule with chain rule
Constant coefficient rule
Derivative of a constant


and so on.

It's imperative you have a list of these rules and that you can recognize when to apply each of them.

Answer 45

thanks man

Answer 46

My pleasure. Altho I need to get off OpenStudy now, i'd be delighted to work with you again. You've written down my e-mail address? I have your yahoo address.

Answer 47

Are you located in a country other than the USA? If so, which one?

Answer 48

i'm from philippines.

Answer 49

so your first language is Tagalog?

Answer 50

i can't add you in facebook

Answer 51

yeah. that's why im not good in english

Answer 52

You can always improve your English.
Let's worry about the Facebook thing later on, ok? for now we can communicate through OpenStudy or e-mail.
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