Master Derivatives with Ease | Solving cos(x-y) = (x)(invcot 2x) + 6^(3x^2 + 2x)

[mesa]

Homework Statement

cos(x-y) = (x)(invcot 2x) + 6^(3x^2 + 2x)

Homework Equations

I'm not sure how to get y to one side, otherwise I think I could handle the derivation

The Attempt at a Solution

So far I have
(cos x)(cos y) + (sin x)(sin y) = (x)(invcot 2x) + 6^(3x^2 + 2x)
I know, not much :)

 

[LCKurtz]

Homework Statement

cos(x-y) = (x)(invcot 2x) + 6^(3x^2 + 2x)

Homework Equations

I'm not sure how to get y to one side, otherwise I think I could handle the derivation

The Attempt at a Solution

So far I have
(cos x)(cos y) + (sin x)(sin y) = (x)(invcot 2x) + 6^(3x^2 + 2x)
I know, not much :)

You could just differentiate it implicitly. Otherwise take the inverse cosine of the original equation as the first step. There will certainly be some domain issues since the left side is between -1 and 1, limiting x values on the right.

 

[HallsofIvy]

Homework Statement

cos(x-y) = (x)(invcot 2x) + 6^(3x^2 + 2x)

Homework Equations

I'm not sure how to get y to one side, otherwise I think I could handle the derivation

The Attempt at a Solution

So far I have
(cos x)(cos y) + (sin x)(sin y) = (x)(invcot 2x) + 6^(3x^2 + 2x)
I know, not much :)

There is no good reason to "reduce" cos(x- y) nor to solve for y. The (partial) derivative of cos(x- y), with respect to x, is -sin(x- y) and the (partial) derivative with respect to y is sin(x- y).
The derivative, with respect to x, of the left side is -sin(x-y)+ sin(x-y)y'. Now differentiate the right side, set those equal, and solve for y'. (IF finding the derivative of y with respect to x is the problem! In the "Problem Statement" section, you give an equation but NO "problem". What, exactly, are you supposed to do with that equation?)