If y(x) = x^x , x gt 0 then y"(2) – 2y'(2) is equal to

The correct Answer is:A

To solve the problem, we need to find y′′(2)−2y′(2) for the function y(x)=xx.

Step 1: Find y′(x)

We start by differentiating y(x)=xx. To do this, we can use logarithmic differentiation.

1. Take the natural logarithm of both sides:
lny=ln(xx)=xlnx

2. Differentiate both sides with respect to x:
1ydydx=lnx+1

3. Multiply through by y:
y′=y(lnx+1)=xx(lnx+1)

Step 2: Find y′(2)

Now we substitute x=2:
y′(2)=22(ln2+1)=4(ln2+1)

Step 3: Find y′′(x)

Next, we differentiate y′(x) to find y′′(x). We will use the product rule:
y′=xx(lnx+1)

Using the product rule:
y′′=ddx(xx)⋅(lnx+1)+xx⋅ddx(lnx+1)

We already found ddx(xx)=xx(lnx+1), and:
ddx(lnx+1)=1x

Thus:
y′′=xx(lnx+1)(lnx+1)+xx⋅1x
=xx(lnx+1)2+xx−1

Step 4: Find y′′(2)

Now we substitute x=2:
y′′(2)=22(ln2+1)2+22−1=4(ln2+1)2+2

Step 5: Calculate y′′(2)−2y′(2)

Now we can calculate:
y′′(2)−2y′(2)=(4(ln2+1)2+2)−2(4(ln2+1))
=4(ln2+1)2+2−8(ln2+1)

Step 6: Simplify the expression

Let’s simplify:
=4(ln2+1)2−8(ln2+1)+2

This is a quadratic in terms of ln2+1:
Let u=ln2+1:
=4u2−8u+2

Step 7: Factor or use the quadratic formula

We can use the quadratic formula:
=4(u2−2u+12)=4((u−1)2−12)

Final Result

Thus, the final answer is:
y′′(2)−2y′(2)=4((ln2+1−1)2−12)=4((ln2)2−12)