Find each derivative in simplest factored form

**Skizye** · December 17th 2009, 09:24 AM

Hi, I'm having some trouble with these problems.

1. $f(x) = 3sin(cos4x)$

2. $f(x) = 5cos(sin(sin(4x^2)))$

3. $f(x) = sin^2(3x^2)$

4. $f(x) = cos^4(sin(cos(3x+2)))$

I'm not really sure what these 2 problems are asking me to do:

5. Write the equation of the tangent line to the curve $f(x) = 3cos2x$ at x = 0.

6. Write the equation of the normal line to the curve $f(x) = tan3x$ at x = pi/12

Any help is appreciated, thanks in advance!

**adkinsjr** · December 17th 2009, 11:19 AM

You aren't suppose to post lists of problems like this. It's againsts the rules of the forum. You should show some of your work so that we know what you're struggling with.

If you're in a calculus class, and you've been assigned these problems, then you should be familiar with the basic rules of differentiation. The functions you listed are all composite trig functions. This means that you need to use the chain rule, and the derivatives of the trig functions in their most basic forms. I'll help you get started with a few of them.

Problem 1

$y=f(x)=3sin(cos(4x))$

Think of it like this. Let $u=cos(4x)$ .

Then $y=3sin(u)$

The derivative can be found by the chain rule.

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

See if you can finish this.

---------------------------------------------------------------

Problem 2

For a function like $y=f(x)=5cos(sin(sin(4x^2)))$

Just apply an extended version of the chain rule.

Let $u=sin(4x^2)$

So you can write:

$y=5cos(sin(u))$

You still have a composite function here, so use another variable:

Let $v=sin(u)$

$y=5cos(v)$

Now just apply the chain rule.

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

See if you can finish.

---------------------------------------------------------------
Problems 5 and 6

For tangent lines, you should be familiar with the fact that the equation of the tangent line to the curve $y=f(x)$ at the point $(a, f(a))$ is $y-f(a) =f'(a)(x-a)$ . This is a basic fact, and should be easy to apply to the functions you were given in problems 5 and 6.

**Skizye** · December 17th 2009, 12:03 PM

Thanks so much! I wasn't aware that this wasn't allowed, I'll make sure to look over the rules before posting again. I have another question, on problem 6, I found the derivative to be

$f'(x) = 3sec^2(3x)$

And I'm trying to plug pi/12 into the equation to find the slope but I'm stuck at this step

$3sec^2(pi/4)$

Where can you go from here? Thanks!

**adkinsjr** · December 17th 2009, 12:58 PM

Originally Posted by Skizye

Thanks so much! I wasn't aware that this wasn't allowed, I'll make sure to look over the rules before posting again. I have another question, on problem 6, I found the derivative to be

$f'(x) = 3sec^2(3x)$

And I'm trying to plug pi/12 into the equation to find the slope but I'm stuck at this step

$3sec^2(pi/4)$

Where can you go from here? Thanks!

Remember the reciprocal identities from trignometry? You can write the slope at $x=\frac{\pi}{12}$ as follows:

$f'\left(\frac{\pi}{12}\right)=3sec^2\left(\frac{\p i}{4}\right)$

$=\frac{3}{cos^2\left(\frac{\pi}{4}\right)}$

It's easy to show that:

$cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$

The square of this is just $\frac{1}{2}$ , so you should find that the slope of the tangent line simplifies to $f'\left(\frac{\pi}{12}\right)=6$ . However, they are asking for the NORMAL line at $x=\frac{\pi}{12}$ . This is the line perpendicular to the tangent line at that point. Perpendicular lines have slopes that are negative reciprocals of each other. So the equation will have the form:

$y-f\left(\frac{\pi}{12}\right)=-\frac{1}{6}\left(x-\frac{\pi}{12}\right)$

$y-tan\left(\frac{\pi}{4}\right)=-\frac{1}{6}\left(x-\frac{\pi}{12}\right)$

You should be able to take it from here. Did that all make sense?

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