# Tricky trig derrivation

1. Oct 22, 2007

### tunabeast

1. The problem statement, all variables and given/known data
Calculate the derivative of tan$$^{2}$$(sin(2x+1)$$^{6}$$)

2. Relevant equations

3. The attempt at a solution
I assume this uses chain rule, by do not see how tan$$^{2}$$ can be derrived.

2. Oct 22, 2007

### neutrino

Can you differentiate (tan(something))2?

3. Oct 22, 2007

### tunabeast

I'm not actually sure i can, this is the first time i'v come accross this kind of differentiation question i'm afraid

4. Oct 22, 2007

### Moridin

If you want to do it in steps, try substituting u = tan(sin(2x + 1)6)

The derivative of u2 where u is a function of something, say x, is

$$2u\frac{du}{dx}$$

Now you have an easier derivative (du/dx). Do this for each part of the chain.

5. Oct 22, 2007

### neutrino

Use the chain rule. For example, consider the function f, which is a function of x, and which in turn is a function of t. If you want to differentiate, say [f(x(t))], use the chain rule...

$$\frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt}.$$

Your problem is more like f(y(x(t))), but the principle remains the same.

6. Oct 22, 2007

### eyehategod

tan^2(sin(2x+1)^6)//let u=2x+1
tan^2(sin(u)^6)//use chain rule
2tan(sin(u)^6)6sin(u)^5u'
2tan(sin(2x+1)^6)(6sin(2x+1)^5)2
24tan(sin(2x+1)^6)(sin(2x+1)^5)
24tan(sin(2x+1)^11)

Last edited: Oct 22, 2007