Question

Find `dy/dx, if y = (cos x)^x + cos^-1sqrtx` is given.

Answer 1

Given `y=(cosx)^x+cos^-1sqrtx`   ...(i)

Let `p = (cosx)^x and q = cos^-1 sqrtx`

for p = (cos x)^x

Taking log both side

log p = x log |(cos x)|

Differentiating both side w.r.t. x

`1/p (dp)/dx = x/cos x (-sinx)+log |(cosx)|`

`(dp)/dx = p[log |(cos x)| - x tan x]`

= (cos x)^x [log |cos x| -x tan x]   ...(ii)

Also q = `cos^-1 sqrtx`   ...(iii)

⇒ `(dq)/dx = (-1)/sqrt(1-(sqrtx)^2)xx1/(2sqrtx)`

`= (-1)/(2 sqrtx - sqrt(1-x))`

Now y = `(cos x)^x + cos^(−1)sqrtx`     ...(from (i)

y = p + q

⇒ `dy/dx = (dp)/dx + (dq)/dx`     ...(from (ii) & (iii))

`dy/dx = (cosx)^x [log |cos x| -x tan x] - 1/(2sqrtx-sqrt(1-x))`