Finding the derivative of the function

1. Jan 24, 2009

meeklobraca

1. The problem statement, all variables and given/known data

f(x) = sin(5x) / cos(x^2)

2. Relevant equations

3. The attempt at a solution

I got 25Cosx-Sin 5xCosx / cos x^3

Would this be the correct answer?

Thanks!

2. Jan 24, 2009

Dick

Nope. Warm up with finding the derivatives of sin(5x) and cos(x^2). What are they? Now apply the quotient rule.

3. Jan 24, 2009

meeklobraca

I have (cos5x)(5)(cosx^2) - (sin5x)(2Cosx) all divided by Cosx^4

Right track?

4. Jan 24, 2009

Dick

It's getting there. But what's the derivative of cos(x^2)? Chain rule. And cos(x^2)^2 is not cos(x^4).

5. Jan 24, 2009

meeklobraca

Ahh I left out the -sin x

Okay so I should have

(cos5x)(5)(Cosx^2) - (sin5x)(2Cosx)(-sinx) / Cosx^4

?

6. Jan 24, 2009

Dick

Why don't you just do the pieces first? Derivative of cos(x^2)? You aren't going to get anywhere without that. Or did you actually mean to write cos(x)^2 in the denominator instead of cos(x^2)?

7. Jan 24, 2009

meeklobraca

The derivative of cos x^2 is -2cosxsinx

8. Jan 24, 2009

Dick

Again, is that (cos(x))^2 or cos(x^2)? They are two different things. Correct if it's the first. Use parentheses!

9. Jan 24, 2009

meeklobraca

Its the 2nd one. And I got what your saying. for Cos(x^2) I got -2xsinx^2

10. Jan 24, 2009

Dick

Write that as -2xsin(x^2) and I'll be happier. And cos(x^2)^2 still isn't cos(x^4). Two other different things.

11. Jan 24, 2009

meeklobraca

Does the same thing apply for Sin (5x) that this is different than Sin 5x?

12. Jan 24, 2009

meeklobraca

Is the denominator 2Cos(x^2)?

13. Jan 24, 2009

Dick

When in doubt use parentheses. sin 5x could mean either sin(5)*x or sin(5x). Better to be safe than sorry.

14. Jan 24, 2009

Dick

Geez. The quotient rule says (u/v)'=(u'*v-v'*u)/v^2. v is cos(x^2). v^2 isn't 2*cos(x^2).

15. Jan 24, 2009

meeklobraca

So I got

(5)Cos (5x)Cos(x^2) - Sin (5x)-2xSin(x^2) / 2Cos (x^2)

(x^2) squared is x^4, what is Cos (x^2)^2 then?

16. Jan 24, 2009

Dick

(cos(x^2))^2 is different from cos((x^2)^2). One is the square of cos(x^2). The other is cos(x^4). They are different. If you don't believe me pick a number and put it into your calculator. Which do you want?

17. Jan 24, 2009

meeklobraca

After doing some review, I got this as a final answer as I dont think I can simplify it anymore.

5Cos(5x)Cos(x^2) + 2x Sin(5x)Sin(x^2) / Cos^2(x^2)

18. Jan 24, 2009

Staff: Mentor

As Dick asked at least a couple of times, use parentheses. Your last answer would be correctly interpreted as
$$5cos(5x)cos(x^2) + \frac{2x sin(5x)Sin(x^2)}{cos^2(x^2)}$$
but I don't think that's what you had in mind.

19. Jan 25, 2009

meeklobraca

I see what your saying. I meant for all of that to be divided by Cos^2(x^2)

20. Jan 25, 2009

Dick

Then I think you've finally got it.