y=tan⁻¹x+cot⁻¹x x ∈ R,Find dy/dx for the given function y wherever defined
Answers
Answered by
0
In the attachment I have answered this problem. The solution of this problem is simple and easy to understand. See the attachment for detailed solution.
Attachments:
Answered by
2
this can be solve with two methods
method 1 :- we know, from inverse trigonometric function,
if we put it in given function, we get,
y = π
now, differentiate with respect to x ,
dy/dx = d(π)/dx = 0 [ as you know , π is constant, so derivative of π = 0]
hence, dy/dx = 0
method 2 :- y = tan^-1x + cot^-1x
differentiate with respect to x,
dy/dx = d(tan^-1x + cot^-1x)/dx
= 1/(1 + x²) + (-1)/(1 + x²)
= 1/(1 + x²) - 1/(1 + x²)
= 0
hence, dy/dx =0
method 1 :- we know, from inverse trigonometric function,
if we put it in given function, we get,
y = π
now, differentiate with respect to x ,
dy/dx = d(π)/dx = 0 [ as you know , π is constant, so derivative of π = 0]
hence, dy/dx = 0
method 2 :- y = tan^-1x + cot^-1x
differentiate with respect to x,
dy/dx = d(tan^-1x + cot^-1x)/dx
= 1/(1 + x²) + (-1)/(1 + x²)
= 1/(1 + x²) - 1/(1 + x²)
= 0
hence, dy/dx =0
rohitkumargupta:
cool:-)
Similar questions