Question

differentiate wrtx (sinx+cox)^1+tannx

Answer 1

Consider y=x^tanx + tanx^sinx

Split it into two parts let A=x^tanx and B= tanx^sinx

(If we have two functions of the variable with one as power of other, then take log on both sides and proceed)

A= x^tanx
implies logA=tanx*logx
differentiating wrt x
implies (1/A)*dA/dx = tanx*(1/x) + (secx)²*logx
implies dA/dx = A*(tanx*(1/x) + (secx)²*logx)
implies dA/dx = (x^tanx)* [(tanx*(1/x) + (secx)²*logx)]
B= tanx^sinx
implies logB= sinx*log(tanx)
differentiating wrt x
implies (1/B)*dB/dx = sinx*(secx)² /tanx+ cosx*tanx
implies dB/dx= (tanx^sinx)* [sinx*(secx)²/tanx + cosx*tanx]
implies dB/dx= (tanx^sinx)* [secx+ sinx]

Now since y=A+B
dy/dx = dA/dx + dB/dx
i.e dy/dx = (x^tanx)* [(tanx*(1/x) + (secx)²*logx)] +
(tanx^sinx)* [secx+ sinx]