If $y=x \cdot \log_e x$, then the value of $\frac{d^2y}{dx^2}$ will be:

(a) $\frac{1}{1+x}$
(b) $\frac{1}{x}$
(c) $\log_e(1+x)$
(d) $1+\log_e x$

Question

If $y=x \cdot \log_e x$, then the value of $\frac{d^2y}{dx^2}$ will be:

(a) $\frac{1}{1+x}$
(b) $\frac{1}{x}$
(c) $\log_e(1+x)$
(d) $1+\log_e x$

Accepted Answer

>

> [!TLDR]
> Differentiate $y=x\log_e x$ twice using the product rule.
>
> Given $y=x\log_e x$.
>
> First derivative (product rule, $u=x$, $v=\log_e x$):
> $\dfrac{dy}{dx}=1\cdot\log_e x + x\cdot\dfrac{1}{x}=\log_e x+1$
> …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.If $y=x \cdot \log_e x$ , then the value of $\frac{d^2y}{dx^2}$ will be:

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Q.If y=x⋅log⁡exy=x \cdot \log_e xy=x⋅loge​x, then the value of d2ydx2\frac{d^2y}{dx^2}dx2d2y​ will be:

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Q.If $y=x \cdot \log_e x$ , then the value of $\frac{d^2y}{dx^2}$ will be:

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