Derivative of tan^2(sin(2x+1)^6) using chain rule | Tricky trig homework

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Homework Help Overview

The discussion revolves around calculating the derivative of the function tan²(sin(2x+1)⁶), focusing on the application of the chain rule in differentiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the differentiation of tan² and question how to apply the chain rule effectively. Some express uncertainty about differentiating this type of function for the first time, while others suggest breaking the problem into manageable parts by substituting variables.

Discussion Status

There is a mix of attempts to clarify the differentiation process, with some participants providing guidance on using substitutions and the chain rule. Multiple interpretations of the problem are being explored, but no consensus has been reached yet.

Contextual Notes

Some participants mention that this is their first encounter with such a differentiation question, indicating a potential lack of familiarity with the concepts involved.

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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.
 
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Can you differentiate (tan(something))2?
 
I'm not actually sure i can, this is the first time i'v come across 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

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

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...

[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.
 
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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