# Derivatives of Trig functions

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• Apr 21st 2010, 10:23 AM
Chokfull
Derivatives of Trig functions
So my calculus book gives me
$(\sin x)'=\cos x$,
and
$(\sec x)'=\sec x \tan x$.
I have to differentiate
$\sin 2t$
and
$3\sec^2 3t$.
So, for the first one, I would take $2t$ to be $x$ and get
$\cos 2t$
but it says it should be
$2\cos 2t$.
I don't understand how this works.
Also for the second one they get
$18\sec^2 3t \tan 3t$,
and using the principle that I learned from the first one it makes sense except that the secant should not be squared.
• Apr 21st 2010, 10:27 AM
Diemo
Both cases are using the chain rule. The chain rule states

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

In the first problem, let u = 2t, in the secont problem, let u = sec(3t). TO solve that you will have to use the chain rule again, letting your y be sec(3t) and your u to be = 3t.
• Apr 22nd 2010, 08:30 AM
Chokfull
Quote:

Originally Posted by Diemo
Both cases are using the chain rule. The chain rule states

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

In the first problem, let u = 2t, in the secont problem, let u = sec(3t). TO solve that you will have to use the chain rule again, letting your y be sec(3t) and your u to be = 3t.

But, every $\sin x=\sin 2a$, if you take $a=\frac {x} {2}$. therefore, the derivative would be $\cos x=2\cos 2a$?
• Apr 22nd 2010, 08:48 AM
Diemo
What? So you let u be 2t. Then you have $\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$. Now, y=sin(u), u=2t.
Hence
$\frac{du}{dt}=2$ and $\frac{dy}{du} = cos(u)$

Hence, $\frac{dy}{dt}= \frac{dy}{du}\frac{du}{dt}=cos(u)*2=2cos(u)=2*cos( 2t)$, where I have resubstituted the u=2t at the end. Now do the same for your second equation, only now you let y= u^2, u = sec(3t), and to work out what the derivitive of sec(3t) is you use the chain rule again, letting u=3t, y=sec(u).
• Apr 22nd 2010, 09:01 AM
AllanCuz
Quote:

Originally Posted by Chokfull
So my calculus book gives me
$(\sin x)'=\cos x$,
and
$(\sec x)'=\sec x \tan x$.
I have to differentiate
$\sin 2t$
and
$3\sec^2 3t$.
So, for the first one, I would take $2t$ to be $x$ and get
$\cos 2t$
but it says it should be
$2\cos 2t$.
I don't understand how this works.
Also for the second one they get
$18\sec^2 3t \tan 3t$,
and using the principle that I learned from the first one it makes sense except that the secant should not be squared.

For the second one,

$3\sec^2 3t$

We need the derivative of the whole and the derivative of the inside. To better see this, let us rewrite

$3\sec^2 3t = 3(sec3t)^2$

So our "outside" is the whole bracket, where we use power rule, and the "inside" is sec3t.

So, the derivative of the whole is

$3(sec3t)^2--->6(sec3t)$

But due to chain rule, we need the derivative of the inside.

$6(sec3t)---->6(sec3t)(sec3t)(tan3t)$

But of course, we can notice that we have another function "3t" that also has a derivative, so we must multiply by this as well.

$6(sec3t)(sec3t)(tan3t)---->6(sec3t)(sec3t)(tan3t)3---->18(sec3t)^2 (tan3t)$
• Apr 23rd 2010, 08:46 AM
Chokfull
Quote:

Originally Posted by Diemo
What? So you let u be 2t. Then you have $\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$. Now, y=sin(u), u=2t.
Hence
$\frac{du}{dt}=2$ and $\frac{dy}{du} = cos(u)$

Hence, $\frac{dy}{dt}= \frac{dy}{du}\frac{du}{dt}=cos(u)*2=2cos(u)=2*cos( 2t)$, where I have resubstituted the u=2t at the end. Now do the same for your second equation, only now you let y= u^2, u = sec(3t), and to work out what the derivitive of sec(3t) is you use the chain rule again, letting u=3t, y=sec(u).

The only problem i see here is that, when differentiating $\sin x$, you can arbitrarily take $z=\frac {x} {2}$, so:

$
\sin x=\sin 2z$

Then, taking the derivative of each side,

$
\cos x=2\cos 2z=2\cos x$

Which solves to $1=2$
• Apr 23rd 2010, 09:11 AM
Diemo
So, you have $y=sin(x)=sin(2z)$, with $x=2z$. Now, you want to defrentiate both sides by x.
LHS:
$\frac{dy}{dx}=\frac{d}{dx}(sin(x))=cos(x)$
RHS:
$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=2cos(2z)* \frac{1}{2}=cos(2z)=cos(x)$

You have to remember what it is you are defrentiating with respect to. For the RHS, you defrentiated with respect to z, for the LHS with respect to x, so of course they are not equal. They are related however, by $\frac{dx}{dz}$
• Apr 24th 2010, 09:05 AM
Chokfull
Quote:

Originally Posted by Diemo
You have to remember what it is you are defrentiating with respect to. For the RHS, you defrentiated with respect to z, for the LHS with respect to x, so of course they are not equal. They are related however, by $\frac{dx}{dz}$

So, you are saying that [tex]F'(x)MATH] does not equal $F'(2z)$, even though $x=2z$?

Hm...I was trying implicit differentiation starting with $\sin x=\sin 2z$, but I suppose that wouldn't work unless one was a function of the other....and so when you multiply the right side by $(2z)'$ you get $2z'$, and when $z=\frac {x} {2}$, $z'$ would be $\frac {1} {2}$.

Also, just a tip, it's spelled "differentiating" lol
• Apr 24th 2010, 09:27 AM
Diemo
Quote:

Originally Posted by Chokfull
So, you are saying that [tex]F'(x)MATH] does not equal $F'(2z)$, even though $x=2z$?

Hm...I was trying implicit differentiation starting with $\sin x=\sin 2z$, but I suppose that wouldn't work unless one was a function of the other....and so when you multiply the right side by $(2z)'$ you get $2z'$, and when $z=\frac {x} {2}$, $z'$ would be $\frac {1} {2}$.

Also, just a tip, it's spelled "differentiating" lol

Yes, exactly.
The text passage contains a discussion about derivatives of trigonometric functions, specifically the differentiation of $\sin 2t$ and $3\sec^2 3t$. The user is confused about the application of the chain rule in these problems and seeks clarification.
multiple_questions_with_answers

1. **question**: how do you differentiate $\sin 2t$?
   - **answer**: the derivative of $\sin 2t$ is $2\cos 2t$.

2. **question**: how do you differentiate $3\sec^2 3t$?
   - **answer**: the derivative of $3\sec^2 3t$ is $18\sec^2 3t \tan 3t$.

the passage includes both the questions and the answers, along with a detailed explanation of the chain rule and its application to these problems.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Derivatives of Trig functions

Printable View

• Apr 21st 2010, 10:23 AM
Chokfull
Derivatives of Trig functions
So my calculus book gives me
$(\sin x)'=\cos x$,
and
$(\sec x)'=\sec x \tan x$.
I have to differentiate
$\sin 2t$
and
$3\sec^2 3t$.
So, for the first one, I would take $2t$ to be $x$ and get
$\cos 2t$
but it says it should be
$2\cos 2t$.
I don't understand how this works.
Also for the second one they get
$18\sec^2 3t \tan 3t$,
and using the principle that I learned from the first one it makes sense except that the secant should not be squared.
• Apr 21st 2010, 10:27 AM
Diemo
Both cases are using the chain rule. The chain rule states

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

In the first problem, let u = 2t, in the secont problem, let u = sec(3t). TO solve that you will have to use the chain rule again, letting your y be sec(3t) and your u to be = 3t.
• Apr 22nd 2010, 08:30 AM
Chokfull
Quote:

Originally Posted by Diemo
Both cases are using the chain rule. The chain rule states

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

In the first problem, let u = 2t, in the secont problem, let u = sec(3t). TO solve that you will have to use the chain rule again, letting your y be sec(3t) and your u to be = 3t.

But, every $\sin x=\sin 2a$, if you take $a=\frac {x} {2}$. therefore, the derivative would be $\cos x=2\cos 2a$?
• Apr 22nd 2010, 08:48 AM
Diemo
What? So you let u be 2t. Then you have $\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$. Now, y=sin(u), u=2t.
Hence
$\frac{du}{dt}=2$ and $\frac{dy}{du} = cos(u)$

Hence, $\frac{dy}{dt}= \frac{dy}{du}\frac{du}{dt}=cos(u)*2=2cos(u)=2*cos( 2t)$, where I have resubstituted the u=2t at the end. Now do the same for your second equation, only now you let y= u^2, u = sec(3t), and to work out what the derivitive of sec(3t) is you use the chain rule again, letting u=3t, y=sec(u).
• Apr 22nd 2010, 09:01 AM
AllanCuz
Quote:

Originally Posted by Chokfull
So my calculus book gives me
$(\sin x)'=\cos x$,
and
$(\sec x)'=\sec x \tan x$.
I have to differentiate
$\sin 2t$
and
$3\sec^2 3t$.
So, for the first one, I would take $2t$ to be $x$ and get
$\cos 2t$
but it says it should be
$2\cos 2t$.
I don't understand how this works.
Also for the second one they get
$18\sec^2 3t \tan 3t$,
and using the principle that I learned from the first one it makes sense except that the secant should not be squared.

For the second one,

$3\sec^2 3t$

We need the derivative of the whole and the derivative of the inside. To better see this, let us rewrite

$3\sec^2 3t = 3(sec3t)^2$

So our "outside" is the whole bracket, where we use power rule, and the "inside" is sec3t.

So, the derivative of the whole is

$3(sec3t)^2--->6(sec3t)$

But due to chain rule, we need the derivative of the inside.

$6(sec3t)---->6(sec3t)(sec3t)(tan3t)$

But of course, we can notice that we have another function "3t" that also has a derivative, so we must multiply by this as well.

$6(sec3t)(sec3t)(tan3t)---->6(sec3t)(sec3t)(tan3t)3---->18(sec3t)^2 (tan3t)$
• Apr 23rd 2010, 08:46 AM
Chokfull
Quote:

Originally Posted by Diemo
What? So you let u be 2t. Then you have $\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$. Now, y=sin(u), u=2t.
Hence
$\frac{du}{dt}=2$ and $\frac{dy}{du} = cos(u)$

Hence, $\frac{dy}{dt}= \frac{dy}{du}\frac{du}{dt}=cos(u)*2=2cos(u)=2*cos( 2t)$, where I have resubstituted the u=2t at the end. Now do the same for your second equation, only now you let y= u^2, u = sec(3t), and to work out what the derivitive of sec(3t) is you use the chain rule again, letting u=3t, y=sec(u).

The only problem i see here is that, when differentiating $\sin x$, you can arbitrarily take $z=\frac {x} {2}$, so:

$
\sin x=\sin 2z$

Then, taking the derivative of each side,

$
\cos x=2\cos 2z=2\cos x$

Which solves to $1=2$
• Apr 23rd 2010, 09:11 AM
Diemo
So, you have $y=sin(x)=sin(2z)$, with $x=2z$. Now, you want to defrentiate both sides by x.
LHS:
$\frac{dy}{dx}=\frac{d}{dx}(sin(x))=cos(x)$
RHS:
$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=2cos(2z)* \frac{1}{2}=cos(2z)=cos(x)$

You have to remember what it is you are defrentiating with respect to. For the RHS, you defrentiated with respect to z, for the LHS with respect to x, so of course they are not equal. They are related however, by $\frac{dx}{dz}$
• Apr 24th 2010, 09:05 AM
Chokfull
Quote:

Originally Posted by Diemo
You have to remember what it is you are defrentiating with respect to. For the RHS, you defrentiated with respect to z, for the LHS with respect to x, so of course they are not equal. They are related however, by $\frac{dx}{dz}$

So, you are saying that [tex]F'(x)MATH] does not equal $F'(2z)$, even though $x=2z$?

Hm...I was trying implicit differentiation starting with $\sin x=\sin 2z$, but I suppose that wouldn't work unless one was a function of the other....and so when you multiply the right side by $(2z)'$ you get $2z'$, and when $z=\frac {x} {2}$, $z'$ would be $\frac {1} {2}$.

Also, just a tip, it's spelled "differentiating" lol
• Apr 24th 2010, 09:27 AM
Diemo
Quote:

Originally Posted by Chokfull
So, you are saying that [tex]F'(x)MATH] does not equal $F'(2z)$, even though $x=2z$?

Hm...I was trying implicit differentiation starting with $\sin x=\sin 2z$, but I suppose that wouldn't work unless one was a function of the other....and so when you multiply the right side by $(2z)'$ you get $2z'$, and when $z=\frac {x} {2}$, $z'$ would be $\frac {1} {2}$.

Also, just a tip, it's spelled "differentiating" lol

Yes, exactly.
user
The text passage contains a discussion about derivatives of trigonometric functions, specifically the differentiation of $\sin 2t$ and $3\sec^2 3t$. The user is confused about the application of the chain rule in these problems and seeks clarification.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

1. **Question**: How do you differentiate $\sin 2t$?
   - **Answer**: The derivative of $\sin 2t$ is $2\cos 2t$.

2. **Question**: How do you differentiate $3\sec^2 3t$?
   - **Answer**: The derivative of $3\sec^2 3t$ is $18\sec^2 3t \tan 3t$.

The passage includes both the questions and the answers, along with a detailed explanation of the chain rule and its application to these problems.