# Homework Help: Trigonometric derivatives

1. Mar 30, 2007

### Ry122

How do I get the derivative of
y=tan4x
The answer in the back of the text book
says (4/cos^2(4X))

2. Mar 30, 2007

### mjsd

remember tan x can be written in terms of sin and cos

3. Mar 30, 2007

### Ry122

yes i know that tanx= sinx/cosx
y=tan4x -> y=sin4x/cos4x
Then what?

4. Mar 30, 2007

### Gib Z

Also use the chain rule, u=4x.

5. Mar 30, 2007

### Ry122

y=sin4x/cos4x
u=4x
u'=4
y=sinu/cosu
y=cosu/-sinu
y=4cosu/-4sinu
y=4cos4x/-4sin4x
What am I doing wrong?

6. Mar 30, 2007

### Gib Z

If you have a function f(x)=u/v then the derivative is f'(x)=$$\frac{u'v-v'u}{v^2}$$. So let u=sin x and v = cos x for this, you can find the derivative of tan x is (sec x)^2, then after that use the chain rule.

7. Mar 30, 2007

### Ry122

Using the quotient rule this is what i get.
u=sin4x
u'=4cos4x
v=cos4x
v'=-4sin4x

(-4sinX)(cos4X)--(4sinX)(sin4X)/(cos4X)^2

8. Mar 30, 2007

### Gib Z

No no, forget about the 4x bit for now, just use the quotient rule with u=sin x and v=cos x to find the derivative of tan x.

One you know that, THEN use the chain rule. Do you know what the chain rule is?

9. Mar 30, 2007

### Moridin

10. Mar 30, 2007

### Ry122

Do you mind showing me the whole working out yourself?
That would help me alot.
Thanks

11. Mar 30, 2007

### ChaoticLlama

The video was more than sufficient.

If you still are unable to complete the problem, seek help from your teacher, peers, or get a tutor.

12. Mar 30, 2007

### f(x)

$$y=tan4x ; u=4x$$
$$\frac{dy}{dx}=\frac{d(tanu)}{du}.\frac{du}{dx}$$
$$\Rightarrow \ \frac{dy}{dx}= sec^2u. 4$$
$$\Rightarrow \ \frac{dy}{dx}=4sec^{2}4x$$

Last edited: Mar 30, 2007
13. Mar 30, 2007

### Ry122

We havent learnt the other ratios yet.
The answer i need to get is in this form
(4/cos^2(4X))
I've finally figured out how to do this i just don't understand how
(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))
What happens to the cos and sin squared in the numerator?

14. Mar 30, 2007

### hotvette

Hint: isn't there a trig identity relating sin2 and cos2?

15. Mar 30, 2007

### Ry122

too easy
thanks guys

16. Mar 30, 2007

### f(x)

i thought you would know $\ sec^2x=\frac{1}{cos^2x}$ :uhh: