Recall from trigonometry that sin (a + a) = sin a *cos a + cos a * sin a. ==> 2sin a * cos a. Now sin2x can be written as sin (x +x) = 2sin x * cos x. Putting back into the question to simplify, we have [2sin x * cos x] / cos x. = 2sin x. Now we find the derivative of 2sin x. Since 2 is a constant we just differentiate sin x. And that will give us cos x i.e 2sin x dx = 2cos x. Therefore the differential of sin2x/ cos x = 2 cos x.

We know that sin 2x = 2* sin x * cos x.

To find the derivative of sin 2x / cos x we first simplify the expression:

sin 2x / cos x

= [2 * sinx * cos x ] / cos x

= 2 sin x

So we get the expression as 2 sin x.

Now we know that the differential of sin x is cos x.

Using that we can find the differential of 2 sin x and it is 2 cos x.

**The required the derivative of ****sin 2x / cos x** **is 2 cos x.**

To find d/dx { sin2x / cosx}

We know d/dx (u(x)/v(x) = [u(x)/v(x)] = {u'(x)* v(x)+u(x)*v'(x)}/(v(x))^2....(1)

(sin2x)' = (cos2x) (2x)' , by chain rule .

So (sin2x)' = 2cos2x.

(cosx)' = -sinx.

Substituting in formula (1), we get:

{(sin2x)/cosx}' = {(2cos2x)cosx - (sin2x)(-sinx)}/cosx)^2

(sin2x/cosx)'= {2cos2x *cosx +sin2x*sinx)}/(cosx)^2.

(sin2x/cosx)' = {2(2cos^2x-1)+2sinx^2*cosx}/(cosx)^2

(sin2x+cosx)' = {4cos^2-2+2+2(1-cos^2x)cosx}/cos^2x

(sin2x/cosx)' = (2cos^2x)/cos^2x = 2cosx.