Find the second-order derivative of the funct | vedclass

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Question

(A) Let $y = x^{3} \log x$. First,we find the first-order derivative using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x^{3}) \cdot \log x + x^{3}

Vedclass · Accepted Answer

$x(5 + 6 \log x)$

Vedclass · Answer

(A) Let $y = x^{3} \log x$. First,we find the first-order derivative using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x^{3}) \cdot \log x + x^{3} \cdot \frac{d}{dx}(\log x)$ $= 3x^{2} \log x + x^{3} \cdot \frac{1}{x}$ $= 3x^{2} \log x + x^{2} = x^{2}(1 + 3 \log x)$. Now,we find the second-order derivative: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}[x^{2}(1 + 3 \log x)]$ $= \frac{d}{dx}(x^{2}) \cdot (1 + 3 \log x) + x^{2} \cdot \frac{d}{dx}(1 + 3 \log x)$ $= 2x(1 + 3 \log x) + x^{2} \cdot \frac{3}{x}$ $= 2x + 6x \log x + 3x$ $= 5x + 6x \log x$ $= x(5 + 6 \log x)$.

Find the second-order derivative of the function $f(x) = x^{3} \log x$.

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