# Thread: Derivatives of Trig functions

1. ## Derivatives of Trig functions

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

2. Both cases are using the chain rule. The chain rule states

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

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.

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

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

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 $\displaystyle \sin x=\sin 2a$, if you take $\displaystyle a=\frac {x} {2}$. therefore, the derivative would be $\displaystyle \cos x=2\cos 2a$?

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

Hence, $\displaystyle \frac{dy}{dt}= \frac{dy}{du}\frac{du}{dt}=cos(u)*2=2cos(u)=2*cos( 2t)$, 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).

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

$\displaystyle 3\sec^2 3t$

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

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

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

So, the derivative of the whole is

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

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

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

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.

$\displaystyle 6(sec3t)(sec3t)(tan3t)---->6(sec3t)(sec3t)(tan3t)3---->18(sec3t)^2 (tan3t)$

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

Hence, $\displaystyle \frac{dy}{dt}= \frac{dy}{du}\frac{du}{dt}=cos(u)*2=2cos(u)=2*cos( 2t)$, 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 $\displaystyle \sin x$, you can arbitrarily take $\displaystyle z=\frac {x} {2}$, so:

$\displaystyle \sin x=\sin 2z$

Then, taking the derivative of each side,

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

Which solves to $\displaystyle 1=2$

7. So, you have $\displaystyle y=sin(x)=sin(2z)$, with $\displaystyle x=2z$. Now, you want to defrentiate both sides by x.
LHS:
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}(sin(x))=cos(x)$
RHS:
$\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=2cos(2z)* \frac{1}{2}=cos(2z)=cos(x)$

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 $\displaystyle \frac{dx}{dz}$

8. 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 $\displaystyle \frac{dx}{dz}$
So, you are saying that [tex]F'(x)MATH] does not equal $\displaystyle F'(2z)$, even though $\displaystyle x=2z$?

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

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

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

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

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