$$(u\cdot v)’=u’\cdot v + u \cdot v’$$
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Deriver uttrykkene
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\begin{align}(x\cdot e^x)'&=1\cdot e^x+x\cdot e^x\\&=e^x(1+x)\end{align}
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\begin{align}(\ln x\cdot x^2)'&=\frac{1}{x}\cdot x^2+\ln x\cdot 2x\\&=x+2x\ln x\\&=x(2\ln x+1)\end{align}
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\begin{align}(e^x \cdot \ln x)'&=e^x \cdot \ln x+e^x\cdot \frac{1}{x}\\&=e^x(\ln x+\frac{1}{x})\end{align}
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\begin{align}(x^2\cdot e^x)'&=2x \cdot e^x+x²\cdot e^x\\ &=x\cdot e^x(2+x)\end{align}
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\begin{align}(2x^3\cdot \ln x)'&=6x² \cdot \ln x + 2x³\cdot \frac{1}{x}\\&=2x²(3\ln x+1)\end{align}
Formler :
$(e^x)’=e^x$
$(\ln x)’=\frac{1}{x}$
Produktregelen :
$(u\cdot v)’=u’\cdot v + u \cdot v’$
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'\[ (2x^2 \cdot \ln(x))' = (2x^2)' \cdot \ln(x) + 2x^2 \cdot (\ln(x))' \]
\[ = 4x \cdot \ln(x) + 2x^2 \cdot \frac{1}{x}\]
\[ = 4x \ln(x) + 2x \]
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\[ ( e^x \cdot (x^2 + 2) )’ = e^x \cdot (x^2 + 2) + e^x \cdot 2x = e^x (x^2 + 2 + 2x) = e^x (x^2 + 2x + 2) \]
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\[ (x\sqrt{x})' = \left(x \cdot x^{1/2}\right)' = \left(x^{3/2}\right)' = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x} \]