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Let 
y=x^(2)ln(x).

(dy)/(dx)=

Let y=x2ln(x) y=x^{2} \ln (x) .\newlinedydx= \frac{d y}{d x}=

Full solution

Q. Let y=x2ln(x) y=x^{2} \ln (x) .\newlinedydx= \frac{d y}{d x}=
  1. Identify Product Rule: We need to use the product rule because yy is the product of two functions of xx, which are x2x^2 and ln(x)\ln(x).
    Product rule: (d/dx)[uv]=u(dv/dx)+v(du/dx)(d/dx)[u*v] = u*(dv/dx) + v*(du/dx)
    Let u=x2u = x^2 and v=ln(x)v = \ln(x).
  2. Differentiate uu: Differentiate uu with respect to xx.dudx=ddx(x2)=2x.\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x.
  3. Differentiate vv: Differentiate vv with respect to xx.dvdx=ddx(ln(x))=1x.\frac{dv}{dx} = \frac{d}{dx}(\ln(x)) = \frac{1}{x}.
  4. Apply Product Rule Formula: Now plug the derivatives and the original functions into the product rule formula.\newline(dydx)=x2(1x)+ln(x)(2x)(\frac{dy}{dx}) = x^{2}*(\frac{1}{x}) + \ln(x)*(2x).
  5. Simplify Expression: Simplify the expression.\newline(dydx)=x+2xln(x)(\frac{dy}{dx}) = x + 2x\ln(x).\newlineOops, made a mistake here, should have been x2x\frac{x^2}{x} which is xx, not just xx.

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