Tricky trig derrivation

  • Thread starter tunabeast
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  • #1
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Homework Statement


Calculate the derivative of tan[tex]^{2}[/tex](sin(2x+1)[tex]^{6}[/tex])


Homework Equations





The Attempt at a Solution


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

Answers and Replies

  • #2
2,076
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Can you differentiate (tan(something))2?
 
  • #3
27
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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
670
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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

[tex]2u\frac{du}{dx}[/tex]

Now you have an easier derivative (du/dx). Do this for each part of the chain.
 
  • #5
2,076
2
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...

[tex]\frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt}.[/tex]

Your problem is more like f(y(x(t))), but the principle remains the same.
 
  • #6
82
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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:

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