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Question
If y = etan x+ (log x)tan x then find dy/dx
Solution
Let y = etan 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]`
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