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Find 
(d)/(dx)(ln(x)e^(x)).
Choose 1 answer:
(A) 
(e^(x))/(x)
(B) 
(e^(x))/(x)+ln(x)
(C) 
(1)/(x)+e^(x)
(D) 
e^(x)((1)/(x)+ln(x))

Find ddx(ln(x)ex) \frac{d}{d x}\left(\ln (x) e^{x}\right) .\newlineChoose 11 answer:\newline(A) exx \frac{e^{x}}{x} \newline(B) exx+ln(x) \frac{e^{x}}{x}+\ln (x) \newline(C) 1x+ex \frac{1}{x}+e^{x} \newline(D) ex(1x+ln(x)) e^{x}\left(\frac{1}{x}+\ln (x)\right)

Full solution

Q. Find ddx(ln(x)ex) \frac{d}{d x}\left(\ln (x) e^{x}\right) .\newlineChoose 11 answer:\newline(A) exx \frac{e^{x}}{x} \newline(B) exx+ln(x) \frac{e^{x}}{x}+\ln (x) \newline(C) 1x+ex \frac{1}{x}+e^{x} \newline(D) ex(1x+ln(x)) e^{x}\left(\frac{1}{x}+\ln (x)\right)
  1. Apply product rule: Use the product rule for derivatives: (ddx)(uv)=uv+uv(\frac{d}{dx})(u\cdot v) = u'v + uv'.
  2. Identify uu and vv: Let u=ln(x)u = \ln(x) and v=exv = e^{x}. Now find uu' and vv'.
  3. Find derivatives: The derivative of ln(x)\ln(x) is 1x\frac{1}{x}, so u=1xu' = \frac{1}{x}.
  4. Use product rule: The derivative of exe^{x} is exe^{x}, so v=exv' = e^{x}.
  5. Simplify expression: Now apply the product rule: (ddx)(ln(x)e(x))=(1x)e(x)+ln(x)e(x).(\frac{d}{dx})(\ln(x)e^{(x)}) = (\frac{1}{x})e^{(x)} + \ln(x)e^{(x)}.
  6. Simplify expression: Now apply the product rule: (ddx)(ln(x)e(x))=(1x)e(x)+ln(x)e(x).(\frac{d}{dx})(\ln(x)e^{(x)}) = (\frac{1}{x})e^{(x)} + \ln(x)e^{(x)}. Simplify the expression: (ddx)(ln(x)e(x))=e(x)x+ln(x)e(x).(\frac{d}{dx})(\ln(x)e^{(x)}) = \frac{e^{(x)}}{x} + \ln(x)e^{(x)}.

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