Prove that the derivative of tan(2x) - cot(2x) equals....

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The derivative of the function f(x) = tan(2x) - cot(2x) is proven to be 16/(1 - cos(8x)) for x in the interval ]0, π/4[. The discussion highlights the importance of proper notation, specifically the use of parentheses in mathematical expressions to avoid misinterpretation. A common error identified was the incorrect transformation of multiplication into addition in the denominator, leading to incorrect results. The correct application of trigonometric identities and careful algebraic manipulation are crucial in deriving the correct derivative.

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  • Study the differentiation of trigonometric functions, focusing on tan(x) and cot(x).
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Let f(x) = tan(2x) - cot(2x) defined on x∈]0,π/4[

Prove that derivative of f(x) is 16/1-cos(8x)

What I did was:

2 * Sin^2(2x) + 2 * Cos^2(2x) / Cos^2(2x) + Sin^2(2x)

If I factor the 2, I reach:

2 * (Sin^2(2x) + Cos^2(2x) / 1+cos(4x)/2 + 1-cos(4x)/22 * 1/ 1 = 2?

What went wrong?
 
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NooDota said:
Let f(x) = tan(2x) - cot(2x) defined on x∈]0,π/4[

Prove that derivative of f(x) is 16/1-cos(8x)
Did you really mean to write ##\frac{16}{1} - cos(8x)##? If not, use parentheses around the terms in the denominator.
NooDota said:
What I did was:

2 * Sin^2(2x) + 2 * Cos^2(2x) / Cos^2(2x) + Sin^2(2x)
Show how you got this. Also, when the numerator or denominator of a fraction has two or more terms, you must put parentheses around them.
NooDota said:
If I factor the 2, I reach:

2 * (Sin^2(2x) + Cos^2(2x) / 1+cos(4x)/2 + 1-cos(4x)/22 * 1/ 1 = 2?

What went wrong?
 
Okay, nvm, solved it.

In the denominator, I turned the multiplication to addition, which is why I got a wrong result.
 

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