\begin{align}(u\cdot v)’=u’\cdot v + u \cdot v’\end{align}

Oppgave 1

\begin{align}(x\cdot e^x)'&=1\cdot e^x+x\cdot e^x\\&=e^x(1+x)\end{align}
\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}
\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}
\begin{align}(x^2\cdot e^x)'&=2x \cdot e^x+x²\cdot e^x\\ &=x\cdot e^x(2+x)\end{align}
\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 :

$(u\cdot v)’=u’\cdot v + u \cdot v’$

Oppgave 2

'\[ (2x^2 \cdot \ln(x))' = (2x^2)' \cdot \ln(x) + 2x^2 \cdot (\ln(x))' \]
\[ = 4x \cdot \ln(x) + 2x^2 \cdot \frac{1}{x}\]
\[ = 4x \ln(x) + 2x \]
\[ ( e^x \cdot (x^2 + 2) )’ = e^x \cdot (x^2 + 2) + e^x \cdot 2x = e^x (x^2 + 2 + 2x) = e^x (x^2 + 2x + 2) \]
\[ (x\sqrt{x})' = \left(x \cdot x^{1/2}\right)' = \left(x^{3/2}\right)' = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x} \]
\begin{align*}f(x)&=x^2\cdot \ln{x}\\f'(x)&=2x\cdot \ln x+x^2\cdot \frac{1}{x}\\&=2x\cdot \ln x+x\\&=x(2\ln x+1)\end{align*}

Oppgave 3

\begin{align*}f(x)&=e^x\cdot (x^2+2)\\u=e^x &, u'=e^x\\v=x^2+2 &, v'=2x\\f'(x)&=e^x(x^2+2)+e^x\cdot 2x\\&=e^x\cdot x^2+e^x\cdot 2+e^x\cdot 2x\\&=e^x(x^2+2x+2)\end{align*}
\begin{align*}f(x)&=(x^2+\pi)e^x\\u=x^2+\pi &, u'=2x\\v=e^x &, v'=e^x\\f'(x)&=2x\cdot e^x+(x^2+\pi)\cdot e^x\\&=e^x(x^2+2x+\pi)\end{align*}
\begin{align*}f(x)&=(x^2+3)x^3\\u=x^2+3 &, u'=2x\\v=x^3 &, v'=3x^2\\f'(x)&=2x\cdot x^3+(x^2+3)\cdot 3x^2\\&=2x^4+3x^4+9x^2\\&=5x^4+9x^2\\&=x^2(5x^2+9)\end{align*}
\begin{align*}f(x)&=x\cdot (x^2+3)\\u=x &, u'=1\\v=x^2+3 &, v'=2x\\f'(x)&=1\cdot (x^2+3)+x\cdot 2x\\&=x^2+3+2x^2\&=3x^2+3\\&=3(x^2+1)\end{align*}