# 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.