Question

Differentiate with respect to x
e^✓sinx+✓cosx​

Answer 1

Step-by-step explanation:

y= sin x./(cos x)^1/2

log y= log sin x -1/2log cos x

1/y.dy/dx=(1/sin x).cos x -1/2(1/cos x).(-sin x)

dy/dx=y[cot x+(1/2).tan x]

dy/dx=(sin x)./(cos x)^1/2[cot x+(1/2).tan x]

dy/dx=(cos x)^1/2+(1/2).(sin x)^2/(cos x)^3/2 answer