Question

diff w.r.t.x y=(25)^log(sec x) - (16)^log (tan x)​

Answer 1

Answer:  0

Step-by-step explanation:

Hey your quest was not clear enough foe me...But i have tried answering from what i understood....So if the answer is wrong pls tell me.

y= $25^{log5 (sec x)}$  -  $16^{log4 (tan x)}$

=> We know 25= 5^2 and 16 =4^2

Substitute,

=>y=  $5^{2. log5(secx)}$ - $4^{2. log 4(tanx)}$

We also know $log (x^{n)}$ = n log x

Apply this

=>y=  $5^{log5 (sec^2x)}$ - $4^{log 4 (tan^2x)}$

Now we know, log a to base a =1

=>y= sec^2x - tan^2x

=> y=1

=> dy/dx = 0

Hope it helps. Pls mark as brainliest