Question

Differentiate y = x^tanx + tanx^sinx w.r.t. x

Answer 1

Let y1 = x^tanx
take log both sides and then differentiate,
y1 w.r.t. x
logy1 = tanx(log(x))
(1/y1).dy1/dx = sec^2(log(x))*1/x
put the value of y1 too now,
dy1/dx = (x^tanx)sec^2(log(x))*1/(x)...................................(1)

let y2 = tanx^sinx
take log both sides and differentiate y2 w.r.t x
logy2 =sinxlog(tanx)
1/y2.dy2/dx = cos(logtanx)*(1/tanx)*sec^2x
dy2/dx = y2*cos(logtanx)*(1/tanx)*sec^2x
dy2/dx = (tanx^sinx)*(cos(logtanx)*(1/tanx)*sec^2x)..........................(2)

y = y1 + y2
or
dy/dx = dy1/dx + dy2/dx
Adding eqns. (1) and (2)
dy/dx = (x^tanx)sec^2(log(x))*1/(x) + (tanx^sinx)*(cos(logtanx)*(1/tanx)*sec^2x)
Further u can reduce it too.