If y(x) = ( x^x )^x,\,x \gt 0 , then d^2x \over dy^2 + 20 at x = 1 is equal to \underline\hspace2cm .

Differentiation PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q41 with Solution

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Correct answer: 16

Given $y(x)=\left(x^x\right)^x, \quad x>0$

Using exponent rules, $\left(x^x\right)^x = x^{x\cdot x}=x^{x^2}$

So, $y=x^{x^2}$

Let $y=x^{x^2}$ Then $\ln y = x^2 \ln x$

Differentiate w.r.t. $x$ : $\frac{1}{y}\frac{dy}{dx} = 2x\ln x + x$

Hence, $\frac{dy}{dx}=y(2x\ln x+x)$

At $x=1$ , $y(1)=1^{1}=1, \quad \ln 1=0$ So, $\left.\frac{dy}{dx}\right|_{x=1}=1\cdot(2\cdot1\cdot0+1)=1$

We have $\frac{dy}{dx}=y(2x\ln x+x)$ Let $f(x)=2x\ln x+x$ Then $\frac{dy}{dx}=yf$

Differentiate again: $\frac{d^2y}{dx^2}=\frac{d}{dx}(yf)=y'f+yf'$ Since $y'=yf$ , $\frac{d^2y}{dx^2}=yf^2+yf'=y(f^2+f')$

Now, $f'(x)=2\ln x+2+1=2\ln x+3$

At $x=1$ , $f(1)=1, \quad f'(1)=3, \quad y(1)=1$ Therefore, $\left.\frac{d^2y}{dx^2}\right|_{x=1}=1(1^2+3)=4$

We need $\frac{d^2x}{dy^2}$

For inverse functions, $\frac{dx}{dy}=\frac{1}{dy/dx}$

Differentiating w.r.t. $y$ ,

=\frac{d}{dx}\left(\frac{1}{y'}\right)\cdot\frac{dx}{dy}$$ This gives the standard formula: $$\frac{d^2x}{dy^2}=-\frac{y''}{(y')^3}\cdot \frac{1}{y'}=-\frac{y''}{(y')^4}$$ So, $$\frac{d^2x}{dy^2}=-\frac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^4}$$ At $x=1$, $$\left.\frac{d^2x}{dy^2}\right|_{x=1}=-\frac{4}{1^4}=-4$$ 5. **Compute the required value** $$\frac{d^2x}{dy^2}+20=-4+20=16$$ Thus, the required integer is $$\boxed{16}$$