If tan inverse(x^2-y^2/x^2+y^2)=a then prove that dy/dx=x/y (1-tana/1+tana).
To prove that dy/dx = x/y * (1 - tan(a) / 1 + tan(a)), we'll start by differentiating both sides of the given equation implicitly with respect to x.
1. Differentiating tan^(-1)((x^2 - y^2) / (x^2 + y^2)) = a with respect to x using the chain rule:
d/dx[tan^(-1)((x^2 - y^2) / (x^2 + y^2))] = d/dx(a)
We will use the identity: d/dx(tan^(-1)(u)) = 1 / (1 + u^2) * (du/dx)
2. On the left-hand side, let's differentiate the inside expression ((x^2 - y^2) / (x^2 + y^2)). We'll call it u for simplicity.
u = (x^2 - y^2) / (x^2 + y^2)
Applying the quotient rule to find du/dx:
du/dx = [(2x)(x^2 + y^2) - (2x)(x^2 - y^2)] / (x^2 + y^2)^2
= [2x (x^2 + y^2 - x^2 + y^2)] / (x^2 + y^2)^2
= [4xy^2] / (x^2 + y^2)^2
Now substituting back into the differentiated equation (1):
1 / (1 + u^2) * (du/dx) = d/dx(a)
Using the identity again, we have:
1 / (1 + [(x^2 - y^2) / (x^2 + y^2)]^2) * ([4xy^2] / (x^2 + y^2)^2) = d/dx(a)
Simplifying the equation further:
[4xy^2] / [(x^2 + y^2)^2 + (x^2 - y^2)^2] = d/dx(a)
To find dy/dx, we can use the relationship dy/dx = (dy/du) * (du/dx) where u = (x^2 - y^2) / (x^2 + y^2).
3. First, let's find dy/du by rearranging the given equation:
tan^(-1)((x^2 - y^2) / (x^2 + y^2)) = a
Applying the tangent function to both sides:
tan(tan^(-1)((x^2 - y^2) / (x^2 + y^2))) = tan(a)
Simplifying the equation:
((x^2 - y^2) / (x^2 + y^2)) = tan(a)
Now, differentiate both sides with respect to u:
d/dx[((x^2 - y^2) / (x^2 + y^2))] = d/dx[tan(a)]
Using the chain rule:
[(2x)(x^2 + y^2) - (2x)(x^2 - y^2)] / (x^2 + y^2)^2 = sec^2(a) * da/dx
By dividing both sides by (2x(x^2+y^2)), we get:
(1 - (x^2 - y^2)/(x^2 + y^2)) / (x/y) = tan(a) * da/dx
(1 - u) / (x/y) = tan(a) * da/dx (Substituting u = (x^2 - y^2) / (x^2 + y^2))
Now we can substitute back into the equation found in step 2:
[4xy^2] / [(x^2 + y^2)^2 + (x^2 - y^2)^2] = (1 - u) / (x/y)
Simplifying further by replacing the denominator:
[4xy^2] / [2x^2 + 2y^2] = (1 - u) / (x/y)
Simplifying again:
2xy^2 / [x^2 + y^2] = (1 - u) / (x/y)
Rearranging the equation:
dy/dx = x/y * (1 - u)
= x/y * (1 - (x^2 - y^2) / (x^2 + y^2))
= x/y * (1 - tan(a))
Finally, we can substitute tan(a) back into the equation:
dy/dx = x/y * (1 - tan(a))
= x/y * (1 - tan(a) / 1 + tan(a))
Therefore, we have proved that dy/dx = x/y * (1 - tan(a) / 1 + tan(a)).
Hint:
write equation as:
tan(a)=(x²-y²)/(x²+y²)
Cross multiply to eliminate denominators:
(x²+y²)tan(a)=(x²-y²)
Implicitly differentiate both sides
(2x+2yy')tan(a) = 2x-2yy'
Group terms and express y'(=dy/dx) in terms of x, y and tan(a) to get the required expression.