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!!