\[\Big(\frac{u}{v}\Big)’=\frac{u’\cdot v-u\cdot v’}{v^2}\]
Oppgave 1
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\begin{align}\Big(\frac{x^2}{e^x}\Big)'&=\frac{2x\cdot e^x-x^2 \cdot e^x}{(e^x)^2}\\&=\frac{e^x(2x-x^2) }{(e^x)^2}\\ &=\frac{x(2+x)}{e^x}\end{align}
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\begin{align}\bigg(\frac{\ln x}{x}\bigg)'&=\frac{\frac{1}{x}\cdot x-\ln x\cdot 1}{x^2}\\ &=\frac{1-\ln x}{x^2}\end{align}
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\begin{align}\bigg(\frac{x}{e^x}\bigg)'&=\frac{1\cdot e^x-x\cdot e^x}{(e^x)²}\\&=\frac{e^x(1-x)}{(e^x)²}\\&=\frac{1-x}{e^x}\end{align}
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\begin{align*}f(x)&= \frac{x^2}{(x+2)}\\u=x^2 &, u'=2x\\v=x+2 &, '=1\\f'(x)&=\frac{2x\cdot(x+2)-x^2\cdot 1}{(x+2)^2}\\&=\frac{2x^2+4x-x^2}{(x+2)^2}\\&=\frac{x^2+4x}{(x+2)^2}\\&=\frac{x(x+4)}{(x+2)^2}\end{align*}
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\begin{align*}f(x)&=\large\frac{2x^3}{e^x}\\u=2x^3 &. u'=6x^2\\v=e^x &. v'=e^x\\f'(x)&=\frac{6x^2\cdot e^x-2x^3\cdot e^x}{(e^x)^2}\\&=\frac{2x^2\cdot e^x(3-x)}{(e^x)^2}\\&=\frac{2x^2(3-x)}{e^x}\end{align*}
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\begin{align*}f(x)&=\large \frac{x+2}{x+1}\\f'(x)&=\frac{1\cdot (x+1)-(x+2)\cdot 1}{(x+1)^2}\\&=\frac{x+1-x-2}{(x+1)^2}\\&=\frac{-1}{(x+1)^2}\end{align*}
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\begin{align*}f(x)&=\large \frac{2x}{\sqrt{x}}\\&=2\cdot x^{1-\frac{1}{2}}\\&=2x^{\frac{1}{2}}\\f'(x)&=2\cdot \frac{1}{2\cdot x^{\frac{1}{2}-1}}\\&=x^{-\frac{1}{2}}\\&=\frac{1}{\sqrt{x}}\\&=\frac{\sqrt{x}}{x}\end{align*}
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15)\begin{align*}f(x)&=\large \frac{x^2+1}{x}\\u=x^2+1 &, u'=2x\\v=x &, v'=1\\f'(x)&=\frac{2x\cdot x-(x^2+1)\cdot 1}{x^2}\\&=\frac{2x^2-x^2-1}{x^2}\\&=\frac{x^2-1}{x^2}\\&=\frac{(x+1)(x-1)}{x^2}\end{align*}
Oppgave 2
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1)\begin{align*}f(x)&=\large \frac{x^2-x-6}{x+2}\\&=\frac{(x-3)(x+2)}{(x+2)}\\&=x-3\\f'(x)&=1\end{align*}
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\begin{align*}f(x)&=\large \frac{e^x}{\ln(x)}\\u=e^x &, u'=e^x\\v=\ln x &, v'=\frac{1}{x}\\f'(x)&=\frac{e^x\cdot \ln x-e^x\cdot \frac{1}{x}}{(\ln x)^2}\\&=\frac{e^x(\ln x-\frac{1}{x})}{(\ln x)^2}\end{align*}
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\begin{align*}f(x)&=\frac{e^x}{x^2}\\f'(x)&=\frac{e^x\cdot x^2-e^x\cdot 2x}{(x^2)^2}\\&=\frac{x\cdot e^x(x-2)}{x^4}\\ &=\frac{e^x(x-2)}{x^3}\end{align*}
Oppgave 3
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\begin{align*}f(x)&=\frac{x^2-1}{x^3}\\u=x^2-1 &, u'=2x\\v=x^3 &, v'=2(x+1)\\f'(x)&=\frac{2x\cdot x^3-(x^2-1)\cdot 3x^2}{(x^3)^2}\\ &=\frac{x^2(2x^2-3x^2+3)}{x^6}\\&=\frac{-x^2+3}{x^4}\end{align*}
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\begin{align*}f(x)&=\frac{x^2}{x-4}\\f'(x)&=\frac{2x(x-4)-x^2\cdot 1}{(x-4)^2}\\&=\frac{2x^2-8x-x^2}{(x-4)^2}\\&=\frac{x^2-8x}{(x-4)^2}\\ &=\frac{x(x-8)}{(x-4)^2}\end{align*}
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\begin{align*}f(x)&=\frac{x+1}{x-2}\\f'(x)&=\frac{1\cdot (x-2)-(x+1)\cdot 1}{(x-2)^2}\\&=\frac{x-2-x-1}{(x-2)^2}\\&=\frac{-3}{(x-2)^2}\end{align*}
Oppgave 4