tan(๐ฅโ๐ฆ)=๐ฆ/1+๐ฅ^2
dy/dx=?
tan(x-y)=y/(1+x^2)
sec^2(x-y)*(1-dy/dx) = [(y+x^2y)(dy/dx)-(y)(2x)] /[(1+x^2)^2]
sec^2(x-y)*(1-dy/dx) = (dy/dx)[1/(1+x^2)] - 2xy/[(1+x^2)^2]
(dy/dx)[sec^2(x-y) + 1/(1+x^2)^2] = sec^2(x-y) + 2xy/[(1+x^2)^2]
dy/dx = [(sec^2(x-y) + 2xy/(1+x^2)^2)] / [sec^2(x-y) + 1/(1+x^2)^2]
Since sec^2(x-y) = 1+tan^2(x-y) = 1+[y^2/(1+x^2)^2], then:
dy/dx = [1+[y^2/(1+x^2)^2] + 2xy/(1+x^2)^2)] / [1+[y^2/(1+x^2)^2] + 1/(1+x^2)^2]
dy/dx = [(1+x^2)^2+y^2+2xy]/[(1+x^2)^2+y^2+1+x^2]
dy/dx = [x^4+2x^2+y^2+2xy+1]/[x^4+3x^2+y^2+2] <-- Final Answer
Assuming the usual carelessness with parentheses, I'll go with
tan(๐ฅ-๐ฆ) = ๐ฆ/(1+๐ฅ^2)
sec^2(x-y) (1 - y') = 1/(1+x^2) y' - 2xy/(1+x^2)^2
y' (-sec^2(x-y) - 1/(1+x^2)) = -sec^2(x-y) -2xy/(1+x^2)^2
y' = (sec^2(x-y) + 2xy/(1+x^2)^2) / (sec^2(x-y) + 1/(1+x^2))
Massaging that won't ever make it look any simpler
This is just an exercise in keeping track of the details in the product and chain rules.
I made substitutions in my answer to make it simpler, but I guess both answers are fine. It depends.
I like yours -- gets rid of the trig stuff.
Good call.
To find dy/dx, we need to differentiate both sides of the given equation with respect to x.
Differentiating tan(x - y) with respect to x involves applying the chain rule. The chain rule states that for a composition of two functions u(v(x)), the derivative is given by du/dx = du/dv * dv/dx.
So, let's break down the differentiation step by step:
1. Start with the given equation: tan(x - y) = y / (1 + x^2).
2. Differentiate both sides with respect to x. On the left side, we differentiate tan(x - y) as follows:
d/dx(tan(x - y)) = d/dx(y / (1 + x^2))
Using the chain rule, we differentiate the outer function (tan) and leave the inner function (x - y) as it is:
sec^2(x - y) * d/dx(x - y) = d/dx(y / (1 + x^2))
The derivative of (x - y) with respect to x is 1, so we have:
sec^2(x - y) = d/dx(y / (1 + x^2))
3. Now, let's differentiate the right side of the equation:
d/dx(y / (1 + x^2)) = d(y) / dx * (1 / (1 + x^2)) + y * d/dx(1 / (1 + x^2))
The derivative of y with respect to x is dy/dx, and the derivative of (1 / (1 + x^2)) is obtained using the quotient rule, which states that for a quotient of two functions u(x) / v(x), the derivative is given by (u'(x) * v(x) - v'(x) * u(x)) / v(x)^2. Applying the quotient rule, we get:
d/dx(1 / (1 + x^2)) = -(2x) / (1 + x^2)^2.
So, the differentiation becomes:
d/dx(y / (1 + x^2)) = dy/dx * (1 / (1 + x^2)) + y * (-(2x) / (1 + x^2)^2)
Simplifying further:
d/dx(y / (1 + x^2)) = dy/dx * (1 / (1 + x^2)) - (2xy) / (1 + x^2)^2
4. Now, let's substitute the simplified derivatives back into the original equation:
sec^2(x - y) = dy/dx * (1 / (1 + x^2)) - (2xy) / (1 + x^2)^2
5. Finally, to find dy/dx, isolate it on one side of the equation:
dy/dx = [sec^2(x - y) + (2xy) / (1 + x^2)^2] * (1 / (1 + x^2))
Therefore, dy/dx is given by: [sec^2(x - y) + (2xy) / (1 + x^2)^2] * (1 / (1 + x^2)).