What am i doing wrong? (differentiate trig function)

I have to differentiate (1/2)[tan(5x)]

My work reads:
let u = tan(5x)
dy/dx = 5sec²(5x)
let y = u/2
du/dx = (u^(-1))/2 = 1/2u

dy/dx = 5sec²(5x).1/2u

=5sec²(5x) / 2(tan(5x))

=5/cos²(5x) / 2sin(5x)/cos(5x)

=5cos(5x) / 2cos²(5x)sin(5x)

=5 / 2cos(5x)sin(5x)

sin(2x)=2sin(x)cos(x) ----- double angle formula
so dy/dx = 5 / sin(10x) = 5cosec(10x)


But the actual answer is 5sec²(5x)/2. What have i done wrong, could someone please tell me what and how to actually solve this? Thank you for any replies

use the chain rule:
y = (1/2)[tan(5x)]
y' = (1/2)[sec^2(5x) * 5]
= 5/2 * sec^2(5x)

You are very welcome and thanks for the BA.

Regards.

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The problem is here:

     Let y = u/2
     du/dx = (u^(-1))/2 = 1/2u

First, you want dy/du since y=u/2 involves only "y" and "u".

Now the exponent is (+1): y = u^(+1) / 2 ≠ u^(0) / 2
So the derivative exponent is (+1-1)=0, not (0-1)=(-1).
And you should know that u^0=1. Put "1" in place of u^(-1) and your work will work.

using chain rule :

if y = (1/2)[tan(5x)] then :

dy/dx = 1/2 sec^2(5x) d/dx (5x)

....= 5/2 sec^2 (5x)