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\title{Calculus en Kansrekenen - assignment 2}
\author{Camil Staps\\\small{s4498062, group Bram}}

\begin{document}

\maketitle
\thispagestyle{fancy}

\begin{enumerate}
    \item \begin{enumerate}
            \item $$\lim\limits_{x\to-\infty}\frac{x^3+2x^2+2}{3x^3+x+4} = \lim\limits_{x\to-\infty}\frac{1+\frac2x+\frac2{x^3}}{3+\frac1x+\frac4{x^3}} = \frac13.$$
            \item $$\lim\limits_{x\to\infty}\frac{3x^2+8}{x+1}=\frac{3x+\frac8x}{1+\frac1x}=\lim\limits_{x\to\infty}3x+\frac8x=\lim\limits_{x\to\infty}3x=\infty.$$
            \item $$\lim\limits_{x\to\infty}\frac{2x+1}{x^2+x}=\lim\limits_{x\to\infty}\frac{\frac2x+\frac1{x^2}}{1+\frac1x}=\frac01=0.$$
            \item \begin{align*}
                        & \lim\limits_{x\to a}\frac{x^n-a^n}{x-a}\\
                    = & \lim\limits_{a+h\to a}\frac{(a+h)^n-a^n}{a+h-a}\qquad\text{with $h=x-a$}\\
                    = & \lim\limits_{h\to0}\frac{(a+h)^n - a^n}{h}\\
                    = & \left(a^n\right)'\\
                    = & n\cdot a^{n-1}.
                \end{align*}
        \end{enumerate}

    \item \begin{enumerate}
            \item $$f'(2) = \lim\limits_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim\limits_{h\to0}\frac0h=0.$$
            \item $$f'(x) = \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} = \lim\limits_{h\to0}\frac{2(x+h)+3-(2x+3)}{h} = \lim\limits_{h\to0}2 = 2.$$
            \item $$f'(3) = \lim\limits_{h\to0}\frac{f(3+h)-f(3)}{h} = \lim\limits_{h\to0}\frac{h^2+6h+9+6+2h+1-(9+6+1)}{h} = \lim\limits_{h\to0}h+8=8.$$
            \item \begin{align*}
                    f'(x) &= \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\\
                          &= \lim\limits_{h\to0}\frac{\frac{5(x+h)-7}{4(x+h)+3} - \frac{5x-7}{4x+3}}{h}\\
                          &= \lim\limits_{h\to0}\frac{\frac{43h}{(4x+4h+3)(4x+3)}}{h}\\
                          &= \lim\limits_{h\to0}\frac{43}{(4x+4h+3)(4x+3)}\\
                          &= \frac{43}{(4x+3)^2}.
                \end{align*}
        \end{enumerate}

    \item \begin{enumerate}
            \item We rewrite this to something of the form $y=ax+b$:
                
                \begin{align*}
                    4y - x + 2(1-\ln2) &= 0\\
                    4y &= x - 2(1-\ln2)\\
                    y &= \frac14x - \frac12(1-\ln2)
                \end{align*}

                For a line $y=ax+b$, $a$ is the slope, so in this case we have the slope $\frac14$.

            \item We equate $f(x)$ and $y$ for $x=2$:

                \begin{align*}
                    \frac14\cdot2 - \frac12(1-\ln2) &= a\ln2 \\
                    \frac12\ln2 &= a\ln2 \\
                    a &= \frac12.
                \end{align*}

            \item For the slope we take the derivative of $f$, i.e. $f'(x) = \frac ax = \frac12a$.

                Then as intercept we take $f(2)-2\cdot f'(2) = a\ln2 - 2\cdot\frac a2 = a\ln2-a$.
                
                The tangent line is then $y=\frac12ax+a\ln2-a$.
        \end{enumerate}

    \item \begin{enumerate}
            \item $f'(x) = 4x^3-6x^2.$
            \item $f'(x) = \frac{(x-7)(2x)-(x^2+5)(1)}{(x-7)^2} = \frac{x+2}{x-7}.$
            \item \begin{align*}
                    f'(x) &= \left(\sin{\sqrt x}\cdot\sin{\sqrt x}\right) \\
                          &= 2\cdot\sin\sqrt x\cdot\left(\sin\sqrt x\right)' \\
                          &= 2\cdot\sin\sqrt x\cdot\cos\sqrt x\cdot\frac12\cdot x^{-\frac12} \\
                          &= \frac12\sin\left(2\sqrt x\right)\cdot\frac1{\sqrt x} \\
                          &= \frac{\sin\left(2\sqrt x\right)}{2\sqrt x}.
                \end{align*}
            \item %todo
            \item $f'(x) = \left(e^{\tan x}\right)' = e(\tan x)\cdot\tan'x = \frac{e(\tan x)}{\cos^2x}$.
            \item $f'(x) = -\frac1{\cos x}\cdot-\sin x=\frac{\sin x}{\tan x}$.
            \item \begin{align*}
                    f'(x) &= \frac1{\sqrt{1-(1-2x)^2}}\cdot-2 \\
                          &= -2\cdot\frac1{\sqrt{4x^2+4x}} \\
                          &= -2\cdot\frac1{2\sqrt{x^2+x}} \\
                          &= -\left(x^2+x\right)^{-\frac12}.
                \end{align*}
            \item %todo
        \end{enumerate}

    \item % todo
\end{enumerate}

\end{document}