$$(f(g(u))’=f’(g(u)\cdot g’(u)$$
Oppgave 1
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\begin{align}f(x)&=x^2+3)^7\\f’(x)&=7(x^2+3)^6\cdot (x^2+3)’\\&=7(x^2+3)^6\cdot 2x\\&=14x(x^2+3)^6\\\end{align}
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\begin{align}f(x)&=e^{2x}\\f’(x) &=e^{2x}\cdot 2\\&= 2e^{2x} \end{align}
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\begin{align} f(x)&=(\ln(x^2 + 2))'\\f’(x) &= \frac{1}{x^2 + 2} \cdot (x^2 + 2)'\\ &= \frac{1}{x^2 + 2} \cdot 2x\\&= \frac{2x}{x^2 + 2} \end{align}
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19)\begin{align*}f(x)&=\ln(x^2)\\&=2\ln x\\f'(x)&=2\cdot \frac{1}{x}\\&=\frac{2}{x}\end{align*}
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\begin{align*}f(x)&=(2x)^3\\f'(x)&=3\cdot (2x)^2\cdot 2\\&=6\cdot 4\cdot x^2\\&=24x^2\end{align*}
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\begin{align*}f(x)&=(x+2)^4\\f'(x)&=4(x+2)^3\end{align*}
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\begin{align*}f(x)&=e^{x^2}\\f'(x)&=e^{x^2}\cdot 2x\\&=2x e^{x^2}\end{align*}
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\begin{align*}f(x)&=\sqrt{2x}\\(\sqrt{u})'&=\frac{1}{2\sqrt{u}}\\u=2x &, u'=2\\f'(x)&=\frac{1}{2\sqrt{2x}}\cdot 2\\&=\frac{1}{\sqrt{2x}}\end{align*}
Oppgave 2
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\begin{align*}f(x)&=(2x^2-4)^3\\f'(x)&=3\cdot (2x^2-4)^2\cdot 4x\\&=12x(2x^2-4)^2\\&=12x(2(x^2-2))^2\\&=12x\cdot 2^2\cdot (x^2-2)^2\\&=12x\cdot 4(x^2-2)^2\\&=48x(x^2-2)^2\end{align*}
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\begin{align*}f(x)&=(x^2-4x+3)^2\\f'(x)&=2(x^2-4x+3)(2x-4)\\&=2\cdot (x-3)(x-1)\cdot 2(x-2)\\&=4(x-1)(x-2)(x-3)\end{align*}
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\begin{align*}f(x)&=\sqrt{x^2-4}\\&=(x^2-4)^{\frac{1}{2}}\\f'(x)&=\frac{1}{2}\cdot (x^2-4)^{\frac{1}{2}-1}\cdot 2x\\&=\frac{2x}{2}\cdot \frac{1}{(x^2-4)^{\frac{1}{2}}}\\&=\frac{x}{\sqrt{x^2-4}}\end{align*}
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\begin{align*}f(x)&=\frac{2x^2}{(3x-2)^2}\\u=2x^2 &,u'=4x\\v=(3x-2)^2 &, v'=2\cdot(3x-2)\cdot 3=6(3x-2)\\f'(x)&=\frac{4x(\mathbf{3x-2})^2-2x^2\cdot 6\mathbf{(3x-2)}}{((\mathbf{3x-2})^2)^2}\\&=\frac{4x(\mathbf{3x-2})(3x-2-3x)}{(\mathbf{3x-2})^4}\\&=\frac{4x\cdot (-2)}{(3x-2)^3}\\&=\frac{-8x}{(3x-2)^3}\end{align*}
Oppgave 3
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\begin{align*}f(x)&=x\cdot \ln(x^2+3)\\f(x)&=1\cdot \ln(x^2+3)+x\cdot \frac{2x}{x^2+3}\\&=\ln(x^2+3)+\frac{2x^2}{x^2+3}\end{align*}
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\begin{align*}f(x)&=(1-x)^3-\ln(1-x)\\f'(x)&=3(1-x)^2(-1)-\frac{1\cdot(-1)}{1-x}\\&=-3(1-x)^2+\frac{1}{1-x}\end{align*}
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\begin{align*}f(x)&=-\ln(1-x)^2\\f'(x)&=\frac{-2\cdot (-1)}{(1-x)}\\&=\frac{2}{1-x}\\&=\frac{-2}{x-1}\end{align*}
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\begin{align*}f(x)&=(3x^2+1)^5\\f'(x)&=5\cdot (3x^2+1)^4\cdot (6x)\\&=30x(3x^2+1)^4\end{align*}
Oppgave 4
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\begin{align*}f(x)&=(x-1)^3\\f'(x)&=3(x-1)^2\end{align*}
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\begin{align*}f(x)&=(1-\sqrt{x})^3\\u=1-\sqrt{x} &, u'=-\frac{1}{2\sqrt{x}}\\f'(x)&=3(1-\sqrt{x})^2\cdot (-\frac{1}{2\sqrt{x}})\\&=-\frac{3(1-\sqrt{x})^2}{2\sqrt{x}}\end{align*}
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\begin{align*}f(x)&=(1+x+x^2)^6\\f'(x)&=6(1+x+x^2)^5\cdot (2x+1)\end{align*}
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\begin{align*}f(x)&=\sqrt{x^2-4}\\u=x^2-4 &; u'=2x\\f'(x)&=\frac{1}{2\sqrt{x^2-4}}\cdot 2x\\&=\frac{x}{\sqrt{x^2-4}}\end{align*}
Oppgave 5
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\begin{align*}f(x)&=\sqrt{2x-x^4}\\u=2x-x^4 &, u'=2-4x^3=2(1-2x^3)\\f'(x)&=\frac{1\cdot 2(1-2x^3)}{2\sqrt{2x-x^4}}\\&=\frac{1-2x^3}{\sqrt{2x-x^4}}\end{align*}
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\begin{align*}f(x)&=\ln(\sqrt{1-x})\\u=\sqrt{1-x} &, u'=\frac{-1}{2\sqrt{1-x}}\\f'(x)&=\frac{1}{\sqrt{1-x}}\cdot \frac{(-1)}{2\sqrt{1-x}}\\ &=\frac{-1}{2(1-x)}\\&=\frac{1}{2(x-1)}\end{align*}
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\begin{align*}f(x)&=e^{3x+1}\\f'(x)&=e^{3x+1}\cdot 3\\&=3e^{3x+1}\end{align*}
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\begin{align*}f(x)&=e^{1-x}\\f'(x)&=e^{1-x}\cdot (-1)\\&=-e^{1-x}\end{align*}
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\begin{align*}f(x)&=\frac{x+1}{(x-2)^4}\\u=x+1 &, u'=1\\v=(x-2)^4 &, v'=4(x-2)^3\\f'(x)&=\frac{1\cdot (x-2)^4-(x+1)\cdot 4(x-2)^3}{((x-2)^4)^2}\\&=\frac{(x-2)^3((x-2)-4(x+1))}{(x-2)^8}\\&=\frac{(x-2)^3(x-2-4x-4))}{(x-2)^8}\\&=\frac{(x-2)^3(-3x-6)}{(x-2)^8}\\ &=\frac{(-3x-6)}{(x-2)^5}\\&=\frac{-3(x+2)}{(x-2)^5}\end{align*}
Oppgave 6
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\begin{align*}f(x)&=x(x+1)^2\\f'(x)&=1\cdot(x+1)^2+x\cdot 2(x+1)\\&=(x+1)(x+1+2x)\\&=(x+1)(3x+1)\end{align*}
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\begin{align*}f(x)&=\sqrt{x^2-9}\\f'(x)&=\frac{1\cdot 2x}{2\sqrt{x^2-9}}\\&=\frac{x}{\sqrt{x^2-9}}\end{align*}
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\begin{align*}f(x)&=(2x-1)^5\\f'(x)&=5\cdot(2x-1)^4\cdot 2\\&=10(2x-1)^4\end{align*}
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\begin{align*}f(x)&=e^{x+3}\\f'(x)&=e^{x+3}\end{align*}
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\begin{align*}f(x)&=5e^{3-8x}\\f'(x)&=5e^{3-8x}\cdot (-8)\\&=-40e^{3-8x}\end{align*}
Oppgave 7