Tricky trig derrivation

Homework Statement

Calculate the derivative of tan$$^{2}$$(sin(2x+1)$$^{6}$$)

The Attempt at a Solution

I assume this uses chain rule, by do not see how tan$$^{2}$$ can be derrived.

Can you differentiate (tan(something))2?

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

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.

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.

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)

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