# Derivative of trig function

1. Sep 25, 2005

### Aresius

Well i've managed to handle these pretty well considering I was absolutely stumped during Limits of trig functions. However on a more difficult homework question I came out with an incredibly huge answer which was far from the real one.

Let me try latex for the first time...

$$\frac {dy} {dx} \frac {\sin(x)\sec(x)} {1+x\tan(x)} = ?$$

I know the answer is (because of the textbook)

$$\frac {1} {(1+x\tan(x))^2}$$

But I came out with a huge answer and i'm stumped. I tried using the product rule for the numerator and then using the quotient rule on the result and the denominator. Keep in mind, we just did those two before doing derivatives of trig functions.

2. Sep 25, 2005

### Integral

Staff Emeritus
first some notation hints:

Perhaps you mean
$$\frac {d} {dx}( \frac {\sin(x)\sec(x)} {1+x\tan(x)}) = ?$$

the y is unnecessary since you are providing a function of x to differentiate.

3. Sep 25, 2005

### grief

can you show what was your solution and how you got there? maybe then we could tell you what step you did wrong.

4. Sep 25, 2005

### Integral

Staff Emeritus
You may wish to consider the definitions of the trig functions to simplify the numerator first. Then apply the quiotient rule,... show us your work.

5. Sep 25, 2005

### Aresius

Actually the way the question was written was find f'(x) but I like Leibniz better.

Well I erased my work because it confused me more than it aided me :rofl: but i'll give it a try.

I applied the product rule to the numerator (sin(x)sec(x)) using the derivatives of basic trig functions, I got this for the numerator.

$$\sin(x)\sec(x)\tan(x) + \sec(x)\cos(x)$$

Looks simple enough but then I applied the quotient rule.

$$\frac {\sin(x)\sec(x)\tan(x) + \sec(x)\cos(x) + \sin(x)\sec(x)\tan^2(x) + \sec(x)\cos(x) - \sin(x)\sec^3(x)} {(1+tan(x))^2}$$

6. Sep 25, 2005

### Integral

Staff Emeritus
From a trig identity stand point what is Sin(x)Sec(x) equivalent to?

7. Sep 25, 2005

### Aresius

Sin(x)/Cos(x)?

Which is Tan(x), therefore i'm an idiot :rofl:

Hold on let me work this out.

Last edited: Sep 25, 2005
8. Sep 25, 2005

### Integral

Staff Emeritus
Nothing any 5yr old couldn't do......

With ten years experiance.

9. Sep 25, 2005

### Aresius

That would make the equation

$$\frac {\tan(x)} {1+\tan(x)}$$

And using the quotient rule I come out with

$$\frac {\sec^2(x)+\sec^2(x)\tan(x)-\tan(x)\sec^2(x)} {(1+\tan(x))^2}$$

That would simplify to

$$\frac {\sec^2(x)} {(1+\tan(x))^2}$$

Which is not the answer in the back of the book. Should I be getting rid of that 1 + before plugging into the quotient rule? I don't think so because that wouldn't have worked in a similar question which I got correct.

10. Sep 25, 2005

### Integral

Staff Emeritus
Isn't the denominator (1 + x tan(x)) ?

11. Sep 25, 2005

### Aresius

Ok I'm going to break something... I swear my brain was here prior to me doing this question.

Would that not give the same result though? Just with an x in front of tanx on the denominator.

Last edited: Sep 25, 2005
12. Sep 25, 2005

### Integral

Staff Emeritus
You will need to use the product rule to differentiate, this will give you an addtional term.

You are close, keep after it.... watch for usefull trig identies.

13. Sep 25, 2005

### Aresius

I differentiated 1+xtan(x) and got xsec^2(x) + tan(x)

Using that I applied the quotient rule and got this

$$\frac {\sec^2(x) + x\tan(x)\sec^2(x) - x\sec^2(x)\tan(x) + \tan^2(x)} {(1-x\tan(x))^2}$$

Which simplifies to

$$\frac {\sec^2(x) + tan^2(x)} {(1-x\tan(x))^2}$$

Now am I assuming correctly that you can apply the pythagorean identity to this and get an answer of 1/1 on the numerator and the denominator staying the same, giving me my correct answer?

14. Sep 25, 2005

### Integral

Staff Emeritus

I get
$$\frac {sec^2(x) - Tan^2(x)} {(1 - x Tan(x))^2}$$

but
$$sec^2(x) - Tan^2(x) =1$$

Last edited: Sep 25, 2005
15. Sep 26, 2005

### Aresius

Yep I got it all right, and got a nice mark for class participation for explaining it to everyone who didn't get it, which just happened to be the entire class :rofl:

Thanks! Now on to Chain rule homework...