Question

y=cot⁻¹ 2x/1/x² x≠±1,Find dy/dx for the given function y wherever defined

Answer 1

I think function is----> $\bf{y=cot^{-1}\frac{2x}{1-x^2}}$

now differentiate with respect to x,

$\frac{dy}{dx}=\frac{d\{cot^{-1}\frac{2x}{1-x^2}\}}{dx}$

= $\frac{-1}{1+\{\frac{2x}{1-x^2}\}^2}\frac{d\{\frac{2x}{1-x^2}\}}{dx}$

= $-\frac{(1-x^2)^2}{(1-x^2)^2+(2x)^2}\frac{(1-x^2)\frac{d(2x)}{dx}-2x\frac{d(1-x^2)}{dx}}{(1-x^2)^2}$

= $-\frac{(1-x^2)^2}{(1+x^2)^2}\frac{2-2x^2+4x^2}{(1-x^2)^2}$

= $-\frac{(1-x^2)^2}{(1+x^2)^2}\frac{2(1+x^2)}{(1-x^2)^2}$

= $-\frac{2}{(1+x^2)}$

Answer 2

HELLO DEAR,






YOUR questions is-----------y = cot-¹ 2x/(1 - x²)

GIVEN:-
Y = cot-¹ {2x/(1 - x²)}

put x= tanα
=> α = tan-¹x

Y = cot-¹ {2tanα/(1 - tan²α)}

Y = cot-¹ (tan2α)

[as tan(π/2 - θ) = cotθ]

Y = cot-¹{cot(π/2 - 2α)}

Y = (π/2 - 2α)

now, dy/dx = dy/dα × dα/dx

=> dy/dx = (0 - 2) × 1/(1 7 x²)

=> dy/dx = -2/(1 + x²)


I HOPE ITS HELP YOU DEAR,
THANKS