Let y=x^(2)ln(x). (dy)/(dx)=

Identify Product Rule: We need to use the product rule because y is the product of two functions of x, which are x^2 and ln(x). Product rule: (d/dx)[u*v] = u*(dv/dx) + v*(du/dx) Let u = x^2 and v = ln(x).Differentiate u: Differentiate u with respect to x. du/dx = d/dx(x^2) = 2x.Differentiate v: Differentiate v with respect to x. dv/dx = d/dx(ln(x)) = 1/x.Apply Product Rule Formula: Now plug the derivatives and the original functions into the product rule formula. (dy/dx) = x^2*(1/x) + ln(x)*(2x).Simplify Expression: Simplify the expression. (dy/dx) = x + 2x*ln(x). Oops, made a mistake here, should have been x^2/x which is x, not just x.

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Let $y=x^{2} \ln (x)$ . $\newline$ $\frac{d y}{d x}=$

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Algebra 2

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Q. Let

y=x^{2} \ln (x)

\newline

\frac{d y}{d x}=

Identify Product Rule: We need to use the product rule because $y$ is the product of two functions of $x$ , which are $x^2$ and $\ln(x)$ .
Product rule: $(d/dx)[u*v] = u*(dv/dx) + v*(du/dx)$
Let $u = x^2$ and $v = \ln(x)$ .
Differentiate $u$ : Differentiate $u$ with respect to $x$ . $\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x.$
Differentiate $v$ : Differentiate $v$ with respect to $x$ . $\frac{dv}{dx} = \frac{d}{dx}(\ln(x)) = \frac{1}{x}.$
Apply Product Rule Formula: Now plug the derivatives and the original functions into the product rule formula. $\newline$ $(\frac{dy}{dx}) = x^{2}*(\frac{1}{x}) + \ln(x)*(2x)$ .
Simplify Expression: Simplify the expression. $\newline$ $(\frac{dy}{dx}) = x + 2x\ln(x)$ . $\newline$ Oops, made a mistake here, should have been $\frac{x^2}{x}$ which is $x$ , not just $x$ .