y= tan(2tan^-1x/2), show that dy/dx= 4(1+y^2)/4+x^2
y = tan(2arctan(x/2))
arctan(y) = 2arctan(x/2)
1/(1+y^2) y' = 4/(4+x^2)
y' = 4(1+y^2)/(4+x^2)
Answer
To find the derivative of the given function y = tan(2tan⁻¹(x/2)), we can use trigonometric and chain rules.
First, let's simplify the expression inside the tangent function. We have tan(2tan⁻¹(x/2)).
The function tan⁻¹(x/2) represents the inverse tangent of (x/2). Let's say the angle θ = tan⁻¹(x/2). This means that tan θ = x/2.
Now, we have y = tan(2θ). We can rewrite this in terms of sin θ and cos θ using the double-angle sine formula:
y = sin(2θ)/cos(2θ).
To find dy/dx, we'll use the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
In our case, f(u) = sin(2u)/cos(2u), and g(x) = θ.
First, let's find the derivative of f(u). We can rewrite f(u) as y = sin(2u) * cos(2u)^(-1).
Using the product rule, dy/du = (d(sin(2u))/du * cos(2u)^(-1)) + (sin(2u) * d(cos(2u)^(-1))/du).
Differentiating sine gives us d(sin(2u))/du = 2cos(2u).
To differentiate cos(2u)^(-1), we can use the chain rule again. Let v = cos(2u). Then, dv/du = -2sin(2u). So, d(cos(2u)^(-1))/du = -1/(v^2) * dv/du = -1/(cos^2(2u)) * -2sin(2u) = 2sin(2u)/(cos^2(2u)).
Now, substituting these derivatives back into dy/du, we have:
dy/du = (2cos(2u) * cos(2u)^(-1)) + (sin(2u) * 2sin(2u)/(cos^2(2u))).
Simplifying this, we get:
dy/du = 2 + 2tan(2u).
Finally, to find dy/dx, we need to compute dy/du * du/dx. Recall that u = θ = tan⁻¹(x/2).
Now, we need to find du/dx. Differentiating θ = tan⁻¹(x/2) with respect to x, we use the chain rule again:
du/dx = (1/(1 + (x/2)^2)) * (1/2) = 1/(2 + x^2/4).
Multiplying dy/du = 2 + 2tan(2u) by du/dx = 1/(2 + x^2/4), we get:
dy/dx = (2 + 2tan(2u)) * (1/(2 + x^2/4)).
Substituting θ = tan⁻¹(x/2) back into the expression, we have:
dy/dx = (2 + 2tan(2tan⁻¹(x/2))) * (1/(2 + x^2/4)).
Simplifying this expression further:
dy/dx = 2/(2 + x^2/4) + 2tan(2tan⁻¹(x/2)) / (2 + x^2/4).
Now, we can use the trigonometric identity tan(2θ) = 2tan(θ) / (1 - tan^2(θ)) to simplify the second term:
dy/dx = 2/(2 + x^2/4) + 2(2tan(tan⁻¹(x/2))) / (1 - (tan(tan⁻¹(x/2)))^2).
Simplifying this further using the fact that tan(tan⁻¹(x/2)) = x/2, we have:
dy/dx = 2/(2 + x^2/4) + 2(2x/2) / (1 - (x/2)^2).
Simplifying the denominator:
dy/dx = 2/(2 + x^2/4) + x/(1 - x^2/4).
Combining the fractions on the right side:
dy/dx = (2 + 4x^2)/(2(2 + x^2/4)) = (4 + 4x^2)/(4 + x^2).
Thus, we have shown that dy/dx = (4 + 4x^2)/(4 + x^2), which is equivalent to 4(1 + y^2)/(4 + x^2).