Find the derivative of x^tanx+(sinx)^cosx w.r.t. x?

1 Answer
May 29, 2018

#(sec^2xtanx)x^(tan x - 1) - cosx(sinx)^(cos x)#

Explanation:

We have,

#d/dx(x^(tan x) + (sin x)^(cos x))#

#= d/dx(x^tan x) + d/dx((sin x)^(cos x))#.........................(i)

Let's find #d/dx(x^tanx)#.

So, #d/dx(x^tanx) = (tanx)x^(tan x - 1) xx sec^2 x = (sec^2xtanx)x^(tan x - 1)# [Chain Rule]

Now, #d/dx((sin x)^(cos x)) = cos x(sin x)^(cos x - 1) xx -sin x = -cosx(sin x)^(cosx - 1+ 1) = -cosx(sinx)^(cos x)# [Here too.]

So, From (i),

#d/dx(x^(tan x) + (sin x)^(cos x)) = d/dx(x^tan x) + d/dx((sin x)^(cos x)) = (sec^2xtanx)x^(tan x - 1) - cosx(sinx)^(cos x)#

And I have no idea how to simplify this further.

Hope this helps.