Question

If y = e^{tan x}+ (log x)^{tan x} then find dy/dx

Answer 1

Let y = e^{tan x}+ (log x)^{tan x}

Put u=e ^tanx and v=(logx)^tanx

y=u+v

`dy/dx=(du)/dx+(dv)/dx`            .......(i)

`u=e^tanx`

Taking logarithm on both sides, we get
log u = tan x.log e = tan x
Differentiating w. r. t. x, we get

`1/u (du)/dx=sec^2x`

`therefore (du)/dx=u.sec^2x`

`therefore (du)/dx=e^(tanx).sec^2x`   ..........(ii)

v = (log x)^{tan x}
Taking logarithm on both sides, we get
log v = tan x.log (log x)

Differentiating w.r.t. x, we get

`1/v (dv)/dx=tanx.1/logx1/x+log(logx)sec^2x`

`(dv)/dx=v[tanx/(xlogx)+log(logx)sec^2x]`

`=(logx)^(tanx)[tanx/(xlogx)+log(logx)sec^2x]` .....(iii)

From (i), (ii) and (iii), we get

`dy/dx=e^tanx.sec^2x+(logx)^(tanx)[tanx/(xlogx)+log(logx)sec^2x]`