Dy/dx ???

Question

2x/sinx 

You then have to prove that the tangent at the point (pi/2 , pi) is y=2x ?

Answer 1

The curve y = 2x/sinx

Differentiating on each side with respect of x.

Apply quotient rule in derivatives d/dx(u/v) = (vu' - uv')/v²

u = 2x, v = sinx

u' = 2, v' = cosx

y' = [sinx(2) - 2x(cosx)]/sin²(x)

y' = 2[sinx - x cosx]/sin²(x)

The tangent point is (pi/2, pi)

Substitute the value x = pi/2 in y'.

y' = 2[sin(pi/2) - (pi/2) cos(pi/2)]/(sin(pi/2))²

y' = 2[1 - 0]/1

y' = 2

This is the slope of tangent line to the curve at (pi/2, pi).

m = 2

To find the tangent line equation, substitute the values of m = 2 and (x, y) = (pi/2, pi) in the slope intercept form of an equation y = mx + b.

pi = 2(pi/2) + b

pi = pi + b

b = 0

Substitute m = 2 and b = 0 in y = mx + b.

Therefore, tangent line is y = 2x.