Differential Function

Question

If !!f(x) = dx^(x^x)/dx!!

Then the value of !!f(1) = ?!!

!!a)" "0!!

!!b)-1!!

!!c)" "1!!

!!d)" "!!None

Solution

In the given differential function we can simplify it by using !!log!! and then differentiating.

Here, !!f(x) = dx^(x^x)/dx!! = dy/dx!!

Lets suppose, !!y = x^(x^x)!!

Now taking !!log!! both sides, we get

!!logy = (x^x).logx!!

Now differentiating both sides, we get

!!(1/y)xxdy/dx = (x^x).(1/x) + (logx). d(x^x)/dx!!

!!(1/y)xxdy/dx = (x^x).(1/x) + (logx). dz/dx!!.......(i)

Lets take !!z=x^x!!

Taking !!log!! both sides, we get

!!logz = x.logx!!

Again differentiating both sides with respect to !!x!!, we get

!!(1/z)xxdz/dx = 1 + logx!!

So, !!dz/dx = (x^x).(1 + logx)!!

putting the value of !!dz/dx!! in equation (i), we get

!!(1/y)xxdy/dx = (x^x).(1/x) + (logx).(x^x).(1 + logx)!!

!!dy/dx = y{(x^x).(1/x) + (logx).(x^x).(1 + logx)}!!

!!f(x) = dy/dx = y{(x^x).(1/x) + (logx).(x^x).(1 + logx)}!!

!!f(x) = x^(x^x){(x^x).(1/x) + (logx).(x^x).(1 + logx)}!!

Now putting !!x=1!!, we get

!!f(1) = 1^(1^1){(1^1).(1/1) + (log1).(1^1).(1 + log1)}!!

!!f(1) = 1.{1.1 + (0).(1).(1 + 0)}!!

!!f(1) = 1!!

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