# Two derivative problems

• Mar 15th 2010, 12:24 PM
Wolvenmoon
Two derivative problems
So, my homework problem is deriving -2*sin(x)*cos(x) as part of using the concavity theorem on -6x^2+cos^2(x). My calculator is saying 2-4(cos(x))^2, but I'm not getting to the same answer.

My other question is working the problem sin^2(cos(4x)), this was a problem on my exam that I got right, but only because I memorized the general pattern of the problem when we got it in class the week before and she just happened to have it on the test. I need to see it worked out step by step, as I get lost in it pretty quickly.

( It works out to -8*sin(4x)*sin(cos(4x)) * cos(cos(4x)) )

Thanks!
• Mar 15th 2010, 12:53 PM
e^(i*pi)
Quote:

Originally Posted by Wolvenmoon
So, my homework problem is deriving -2*sin(x)*cos(x) as part of using the concavity theorem on -6x^2+cos^2(x). My calculator is saying 2-4(cos(x))^2, but I'm not getting to the same answer.

My other question is working the problem sin^2(cos(4x)), this was a problem on my exam that I got right, but only because I memorized the general pattern of the problem when we got it in class the week before and she just happened to have it on the test. I need to see it worked out step by step, as I get lost in it pretty quickly.

( It works out to -8*sin(4x)*sin(cos(4x)) * cos(cos(4x)) )

Thanks!

Use an identity:

$-2\sin(x)\cos(x) = -\sin(2x)$

$-\frac{d}{dx} \sin(2x) = -2\cos(2x)$

$\cos(2x) = 2\cos^2(x)-1$ so $-2\cos(2x) = -4\cos^2(x)+2 = 2-4\cos^2(x)$ which is the calculator's reply.

=================================

You can also use the product rule

$-2 \frac{d}{dx} = \sin(x)\cos(x)$

$u = \sin(x) \: \: \rightarrow \: \: u' = \cos(x)$

$v = \cos(x) \: \: \rightarrow \: \: v' = -\sin(x)$

By the product rule

$-2(-\sin^2(x)+\cos^2(x)) = 2\sin^2(x)-2\cos^2(x)$

However, given that $\sin^2(x) = 1-\cos^2(x)$

$2-2\cos^2(x)-2\cos^2(x) = 2-4\cos^2(x)$
• Mar 31st 2010, 04:25 AM
mathemagister
Quote:

Originally Posted by Wolvenmoon
So, my homework problem is deriving -2*sin(x)*cos(x) as part of using the concavity theorem on -6x^2+cos^2(x). My calculator is saying 2-4(cos(x))^2, but I'm not getting to the same answer.

My other question is working the problem sin^2(cos(4x)), this was a problem on my exam that I got right, but only because I memorized the general pattern of the problem when we got it in class the week before and she just happened to have it on the test. I need to see it worked out step by step, as I get lost in it pretty quickly.

( It works out to -8*sin(4x)*sin(cos(4x)) * cos(cos(4x)) )

Thanks!

Use:

The Chain Rule

$\boxed{\frac{d}{dx}f\left[g(x)\right] = f'\left[g(x)\right]\cdot g'(x)}$

Quote:

Originally Posted by Wolvenmoon
I memorized the general pattern of the problem

If this is the "general pattern" you memorized, you did the right thing.

In this case you want to differentiate $\sin^2[\cos(4x)]$

Using the chain rule: $f(x) = \sin^2x, \ \ g(x) = \cos(4x)$

Note that f(x) is itself a combo function with the $\sin{x}$ trapped inside the $(\sin{x})^2$

So the derivative of $(\sin{x})^2$ is $f'(x) = 2\sin{x} \cdot \frac{d}{dx} \sin{x} = 2\sin{x}\cos{x}$

A trig identity says that $2\sin{x}\cos{x} = \sin{(2x)}$

So $f'(x) = \sin{(2x)}$

Notice the g(x) is also a combo with the 4x trapped inside the $\cos{(4x)}$.

So the derivative of $\cos{(4x)}$ is $g'(x) = -\sin{(4x)} \cdot 4 = -4\sin{(4x)}$

Now using the master chain rule:

The derivative of $sin^2[cos(4x)]$ is $f'[g(x)]\cdot g'(x)$

Substitute g'(x) and substitute g(x) for x in the f'(x) calculation:

$\sin(2cos(4x))\cdot -4sin(4x) = -4\sin[2\cos{(4x)}] \sin{(4x)}$

Phew! (Whew) And that's the answer.

NOTE: This answer is a bit more simplified that the one you got because I used the trig identity $2\sin{x}\cos{x} = \sin{(2x)}$. You can check that they are both the same function, by graphing them or randomly checking points such as x=3. I have checked and saw that they are the same, so my answer is also correct.

Hope that helps :)

Mathemagister