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Question

Mathematics Question on Derivatives

If y(x)=xx,x>0y(x)=x^x, x > 0, then y′′(2)−2y′(2)y^{\prime \prime}(2)-2 y^{\prime}(2) is equal to :

A

4(log⁡e2)2−24\left(\log _{ e } 2\right)^2-2

B

8log⁡e2−28 \log _{ e } 2-2

C

4(log⁡e2)2+24\left(\log _e 2\right)^2+2

D

4log⁡e2+24 \log _{ e } 2+2

Answer

4(log⁡e2)2−24\left(\log _{ e } 2\right)^2-2

Explanation

Solution

y′=xx
y′=xx(1+ℓnx)
y′′=xx(1+ℓnx)2+xx⋅x1​
y′′(2)=4(1+ℓn2)2+2
y′(2)=4(1+ℓn2)
y′′(2)−2y′(2)=4(1+ℓn2)2+2−8(1+ℓn2)
=4(1+ℓn2)[1+ℓn2−2]+2
=4(ℓn2)2−1)+2
=4(ℓn2)2−2