Derivatives of Trig Functions

[suicoted]

I don't know how to differentiate y = (cot^2 (2x)) /(1 + x^2)
*** only the 2 is superscript after the cot.

I don't know what I did wrong, but I did not get the answer. Could someone help me with the method, thx?
I used the quotient rule and got stuck.

4(cot 2x)(1 + x^2)-(cot 2x)^2 (2x)/(1+x^2)^2


The final answer in the textbook is : -2cot2x[2(1+x^2)csc^2 2x + x cot 2x]/ (1 + x^2)^2

 

[ChaoticLlama]

Hello.

The first thing to keep in mind is chain rule, it is very important when differentiating trig functions.

The derivative of cot(x) is -csc²(x)

The derivative of cot(2x) is -2csc²(x)

The derivative of cot²(2x) is -4cot(2x)csc²(2x)

y = cot²(2x) / (1 + x²)

Yes, quotient rule is the way to go.

y' = [(1 + x²) * d(cot²(2x))/dx - cot²(2x) * d(1 + x²)/dx] / (1 + x²)²

y' = [(1 + x²) * -4cot(2x)csc²(2x) - cot²(2x) * 2x] / (1 + x²)²

y' = [-4(1 + x²)cot(2x)csc²(2x) - 2xcot²(2x)] / (1 + x²)²

Now, just factor.

y' = {-2cot(2x)[2(1 + x²)csc²(2x) + xcot(2x)]} / (1 + x²)²

 

[suicoted]

The derivative of cot²(2x) is -4cot(2x)csc²(2x) , how?

if the derivative of cot x is -csc^2 x, where did the -4 cot come from? I understand the method used to solve the problem, but I'm just not seeing how the derivative of cot²(2x) is -4cot(2x)csc²(2x). By the way, thanks for your help. Most things are clarified now.

 

[tkhunny]

The derivative of cot²(2x) is -4cot(2x)csc²(2x) , how?

The Chain Rule is a beautiful thing. You don't have just the cotangent. The derivative is with respect to 'x'. You must recognize ALL complications. The cotangent is only one of three on this problem.

First, the exponent of 2. Use the polynomial rule to get the first piece: 2*cot(2x)

Second, the cotangent. Use what you know of the cotangent to get the second piece: -csc^{2}(x)

Third, the factor of 2 inside the cotangent argument. Use the polynomial rule to get the last piece: 2

Put it all together: 2*cot(2x)*(-csc^{2}(x))*2

There's a little algebra left, but I'll let you do that.

 

[suicoted]

The derivative of cot²(2x) is -4cot(2x)csc²(2x) , how?

The Chain Rule is a beautiful thing. You don't have just the cotangent. The derivative is with respect to 'x'. You must recognize ALL complications. The cotangent is only one of three on this problem.

First, the exponent of 2. Use the polynomial rule to get the first piece: 2*cot(2x)

Second, the cotangent. Use what you know of the cotangent to get the second piece: -csc^{2}(x)

Third, the factor of 2 inside the cotangent argument. Use the polynomial rule to get the last piece: 2

Put it all together: 2*cot(2x)*(-csc^{2}(x))*2

There's a little algebra left, but I'll let you do that.

Thx so much you guys, both of you! Calculus can be 'fun' once you get the hang of it . Yeah, the chain rule is the best rule... that is, if I don't mess it up. :roll: