Question

Differentiate the following w.r.t. x :

e^tanx + (logx)^tanx 

Answer 1

Let y = e^tanx + (logx)^tanx 
Put u = (logx)^tanx  
∴ log u = log (log x)^tanx = (tan x).(log log x)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[(tan x).(log logx)]`

= `(tanx)."d"/"dx"(log log x) + (log log x)."d"/"dx"(tan x)`

= `tanx xx 1/logx."d"/"dx"(logx) + (log log x)(sec^2x)`

= `tanx xx 1/logx xx 1/x + (log log x)(sec^2x)`

∴ `"du"/"dx" = u[tanx/(xlogx) + (loglogx)(sec^2x)]`

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

Now, y = e^tanx + u
∴ `"dy"/"dx" = "d"/"dx"(e^tanx) + "du"/"dx"`

= `e^(tanx)."d"/"dx"(tanx) + "du"/"dx"`

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