JEE Main 2022 — Mathematics Calculus

Given,

$$y(x)={({x}^{x})}^{x}$$y(x)=(xx)x

Taking ${\mathrm{log}}_{e}$loge​ both side

$$\mathrm{ln}y(x)={x}^{2}\cdot \mathrm{ln}x$$lny(x)=x2⋅lnx

Now differentiating both side w.r.t $x$x we get,

$$\frac{1}{y(x)}\cdot {y}^{'}(x)=\frac{{x}^{2}}{x}+2x\cdot \mathrm{ln}x$$y(x)1​⋅y′(x)=xx2​+2x⋅lnx

${y}^{'}(x)=y(x)[x+2x\mathrm{ln}x]$y′(x)=y(x)[x+2xlnx] .......(i)

Given $y(1)=1$y(1)=1, so ${y}^{'}(1)=1$y′(1)=1

Now rewriting equation (i) again we get,

$$\frac{dx}{dy}=\frac{1}{{x}^{{x}^{2}+1}(1+2\mathrm{ln}x)}$$dydx​=xx2+1(1+2lnx)1​

Now $\frac{{d}^{2}x}{d{y}^{2}}=\frac{d}{dx}({({x}^{{x}^{2}+1}(1+2\mathrm{ln}x))}^{-1})\frac{dx}{dy}$dy2d2x​=dxd​((xx2+1(1+2lnx))−1)dydx​

$$\Rightarrow \frac{{d}^{2}x}{d{y}^{2}}=\frac{-{x}^{{x}^{2}}(1+2\mathrm{ln}x)({x}^{2}+3+2{x}^{2}\mathrm{ln}x)}{{({x}^{{x}^{2}}(1+2\mathrm{ln}x))}^{3}}\times 1$$⇒dy2d2x​=(xx2(1+2lnx))3−xx2(1+2lnx)(x2+3+2x2lnx)​×1

$$\Rightarrow {(\frac{{d}^{2}x}{d{y}^{2}})}_{x=1}=-4$$⇒(dy2d2x​)x=1​=−4

So, ${(\frac{{d}^{2}x}{d{y}^{2}})}_{x=1}+20=-4+20=16$(dy2d2x​)x=1​+20=−4+20=16