Question

Find `dy/dx`, if y = (sin x)^{tan x} – x^{log x}.

Answer 1

Let y = (sin x)^{tan x} – x^{log x}

Let u = (sin x)^{tan x} and v = x^{log x}

∴ y = u − v, where u and v are differentiable functions of x.

`dy/dx = (du)/dx - (dv)/dx`   ....(I)

Now, u = (sin x)^{tan x}, taking log of both the sides we get,

log u = log (sin x)^{tan x}

∴ log u = tan x log (sin x)

Differentiate w. r. t. x.

`d/dx (logu) = d/dx [tanx log (sinx)]`

`1/u (du)/dx = tanx d/dx[log(sinx)] + log(sinx) d/dx(tanx)`

= `tanx. 1/sinx.d/dx.(sinx) + log(sinx).(sec^2)`

`(du)/dx = u[tanx. 1/sinx.(cosx) + sec^2x.log(sinx)]`

`(du)/dx = (sin x)^(tan x) [tanx.cotx + sec^2x.log(sinx)]`

`(du)/dx = (sin x)^tan x [1 + sec^2x.log(sin x)]`   ....(II)

And v = x^{log x}

Taking log on both the sides we get,

log v = log (x^{log x})

log v = log x log x = (log x)²

Differentiate w. r. t. x.

`d/dx(logv) = d/dx[(logx)^2]`

`1/v (dv)/dx = 2logx d/dx(logx)`

`(dv)/dx = u[(2logx)/x] = (2x^logx logx)/x`  ....(III)

Substituting (II) and (III) in (I) we get,

`dy/dx = (sinx)^tanx [1 + sec^2x.log(sinx)] - (2x^logx logx)/x`