If y(x) = x^x, x 0,then y^{prime prime}(2) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $y(x) = x^x$. Taking the natural logarithm on both sides,$\ln y = x \ln x$.Differentiating with respect to $x$,$\frac{1}{y} y^{\prime} = 1 \

Vedclass · Accepted Answer

$4(\log_e 2)^2 - 2$

Vedclass · Answer

(C) Given $y(x) = x^x$. Taking the natural logarithm on both sides,$\ln y = x \ln x$.Differentiating with respect to $x$,$\frac{1}{y} y^{\prime} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.Thus,$y^{\prime} = y(1 + \ln x) = x^x(1 + \ln x)$.Now,differentiate $y^{\prime}$ with respect to $x$ using the product rule:$y^{\prime \prime} = \frac{d}{dx}[x^x] \cdot (1 + \ln x) + x^x \cdot \frac{d}{dx}[1 + \ln x]$$y^{\prime \prime} = x^x(1 + \ln x)(1 + \ln x) + x^x \cdot \frac{1}{x} = x^x(1 + \ln x)^2 + x^{x-1}$.At $x = 2$:$y^{\prime}(2) = 2^2(1 + \ln 2) = 4(1 + \ln 2)$.$y^{\prime \prime}(2) = 2^2(1 + \ln 2)^2 + 2^{2-1} = 4(1 + \ln 2)^2 + 2$.Now calculate $y^{\prime \prime}(2) - 2y^{\prime}(2)$:$= 4(1 + \ln 2)^2 + 2 - 2[4(1 + \ln 2)]$$= 4(1 + 2\ln 2 + (\ln 2)^2) + 2 - 8 - 8\ln 2$$= 4 + 8\ln 2 + 4(\ln 2)^2 + 2 - 8 - 8\ln 2$$= 4(\ln 2)^2 - 2$.

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

Similar Questions

Differentiate $\sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}}$ with respect to $x$.

If $f(\theta) = \cos \theta_1 \cdot \cos \theta_2 \cdot \cos \theta_3 \cdots \cos \theta_n$,then $\tan \theta_1 + \tan \theta_2 + \tan \theta_3 + \cdots + \tan \theta_n =$

If $h(x) = x^{x^x}$, then at $x = 1$, $\frac{h^{\prime}(x)}{h(x)}$ is equal to

If $f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}$ and $\frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+1}+\operatorname{coth} x+g(x)$, then $g(x)=$

Differentiate the function with respect to $x$: $(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module