Purpose of the project: Struggle through a difficult but important calculus problem.

Each of the following is a difficult definition of the derivative problem. Your group will be assigned one of the following, and then you can present the solution to the class. In each case, the ``stuff in the example box" is not your problem, but look at it and hopefully it will help with your problem.

1. Let [latex]f(x) = x^4[/latex]. Using the definition of the derivative, find [latex]f'(x)[/latex].
   Simplify [latex](x + 1)^4[/latex].We can do this problem by splitting it up into [latex](x + 1)^2 (x + 1)^2[/latex]. We know [latex](x + 1)^2 = x^2 + 2x + 1[/latex] -- that means

   [latex](x + 1)^4 = (x^2 + 2x + 1)(x^2 + 2x + 1)[/latex]

   To solve from here, we need to multiply every term of [latex](x^2 + 2x + 1)[/latex] by [latex]x^2[/latex], then every term by [latex]2x[/latex], then every term by [latex]1[/latex], and add it all up. Here we go: \begin{align*}
   (x + 1)^4 & = (x^2 + 2x + 1)({\color{green} x^2} + {\color{red} 2x} + {\color{blue} 1}) \\
   & = (x^2 + 2x + 1){\color{green}x^2} + (x^2 + 2x + 1){\color{red}2x} + (x^2 + 2x + 1){\color{blue}1} \\
   & = {\color{green} x^4 + 2x^3 + x^2} + {\color{red}2x^3 + 4x^2 + 2x} + {\color{blue}x^2 + 2x + 1} \\
   & = x^4 + 4x^3 + 6x^2 + 4x + 1.
   \end{align*}

2. Let [latex]f(x) = \sqrt{x}[/latex]. Using the definition of the derivative, find [latex]f'(x)[/latex] (which can also be written [latex]\frac{d}{dx} f(x)[/latex] or [latex]\frac{df}{dx}[/latex]).
   Rationalize the numerator of [latex]\frac{\sqrt{x + 1} - \sqrt{x}}{x}[/latex].

   To ``rationalize the numerator", the trick is to multiply top and bottom by what is known as the conjugate: it is the same as the numerator, only the sign is flipped so that subtraction becomes addition or vice versa. In this case, the conjugate is [latex]\sqrt{x + 1} + \sqrt{x}[/latex]. We see
   \begin{align*}
   \frac{\sqrt{x + 1} - \sqrt{x}}{x} & = \frac{(\sqrt{x + 1} - \sqrt{x})}{(x)} \cdot \frac{(\sqrt{x + 1} + \sqrt{x})}{(\sqrt{x + 1} + \sqrt{x})}\\
   & = \frac{(\sqrt{x + 1} - \sqrt{x})(\sqrt{x + 1} + \sqrt{x})}{(x)(\sqrt{x + 1} + \sqrt{x})} \\
   & = \frac{(\sqrt{x + 1})^2 - (\sqrt{x})^2}{x(\sqrt{x + 1} + \sqrt{x})} \\
   & = \frac{x+1 - x}{x(\sqrt{x + 1} + \sqrt{x})} \\
   & = \frac{1}{x(\sqrt{x + 1} + \sqrt{x})}
   \end{align*}
   The numerator is now rationalized.

3. Let [latex]f(x) = \sin(x)[/latex]. Using the definition of the derivative, find [latex]f'(x)[/latex] (which can also be written [latex]\frac{d}{dx} f(x)[/latex] or [latex]\frac{df}{dx}[/latex]). There are three facts we need to compute this derivative:
   • [latex]\sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)[/latex]
   • [latex]\lim_{h \to 0} \frac{\sin(h)}{h} = 1[/latex]
   • [latex]\lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0[/latex]
   Find [latex]\lim_{h \to 0} \cfrac{\sin(y + h) - \sin(y) \cos(h)}{h}[/latex].

   In this example, we use the first fact listed above to write [latex]\sin(y + h) = \sin(y) \cos(h) + \cos(y) \sin(h)[/latex]. We have the original problem is equal to
   \begin{align*}
   & = \lim_{h \to 0} \frac{\sin(y) \cos(h) + \cos(y) \sin(h)- \sin(y) \cos(h)}{h} \\
   & = \lim_{h \to 0} \frac{\cos(y) \sin(h)}{h} \\
   & = \lim_{h \to 0} \frac{\sin(h)}{h} \cdot \cos(y) \\
   & = 1 \cdot \cos(y) \\
   & = \cos(y) \end{align*}
   Notice towards the end we used [latex]\lim_{h \to 0} \frac{\sin(h)}{h} = 1[/latex].

4. Let [latex]f(x) = e^x[/latex]. Using the definition of the derivative, find [latex]f'(x)[/latex] (which can also be written [latex]\frac{d}{dx} f(x)[/latex] or [latex]\frac{df}{dx}[/latex]). Use the limit we found on the homework yesterday: [latex]\lim_{h \to 0} \frac{e^h - 1}{h} = 1[/latex].
   Helpful example: Simplify [latex]\frac{e^{x + 1} + e^1}{e^x + 1}[/latex].

   Recall that [latex]e^{x + 1} = e^x e^1[/latex]. Hence this becomes
   \begin{align*}
   \frac{e^{x + 1} + e^1}{e^x + 1} & = \frac{e^x e^1 + e^1}{e^x + 1} \\
   & = \frac{e^1(e^x + 1)}{e^x + 1} \\
   & = e^1 = e
   \end{align*}

5. Let [latex]f(x) = \frac{1}{x}[/latex]. Using the definition of the derivative, find [latex]f'(x)[/latex] (which can also be written [latex]\frac{d}{dx} f(x)[/latex] or [latex]\frac{df}{dx}[/latex]).
   Simplify [latex]\cfrac{\cfrac{1}{x + 1} - \cfrac{1}{x}}{x}[/latex].

   In this example, we want to clear the denominators from this tricky double fraction. To do this, we will multiply by both denominators: [latex]x(x + 1)[/latex].
   \begin{align*}
   \cfrac{\cfrac{1}{x + 1} - \cfrac{1}{x}}{x} & = \cfrac{\left(\cfrac{1}{x + 1} - \cfrac{1}{x} \right)}{x} \cdot \frac{x(x + 1)}{x(x + 1)} \\
   & = \cfrac{\left(\cfrac{1}{x + 1} - \cfrac{1}{x} \right)x(x + 1)}{(x)(x)(x + 1)} \\
   & = \cfrac{\left(\cfrac{x(x + 1)}{x + 1} - \cfrac{x(x + 1)}{x} \right)}{x^2(x + 1)} \\
   & = \cfrac{x - (x + 1)}{x^2(x + 1)} \\
   & = \cfrac{-1}{x^2(x + 1)}
   \end{align*}

6. Let [latex]f(x) = \ln(x)[/latex]. Using the definition of the derivative, find [latex]f'(x)[/latex] (which can also be written [latex]\frac{d}{dx} f(x)[/latex] or [latex]\frac{df}{dx}[/latex]). We will need the following facts:
   • [latex]e^x = \lim_{h \to 0} \left(1 + hx \right)^{1/h}[/latex]
   • [latex]\ln(A) - \ln(B) = \ln(A/B)[/latex]
   • [latex]n \ln(A) = \ln(A^n)[/latex]
   • [latex]\ln(e^x) = x[/latex]
   (Example 1): Rewrite [latex]\frac{\ln(x+5) - \ln(x)}{2}[/latex] as the natural log of a single quantity.

   We need to use some log rules to simplify this. First we use [latex]\ln(A) - \ln(B) = \ln(A/B)[/latex].
   \begin{align*}
   \frac{\ln(x+5) - \ln(x)}{2} & = \frac{\ln\left(\frac{x+5}{x}\right)}{2} \\
   & = \frac{\ln\left(1 + \frac{5}{x} \right)}{2}
   \end{align*}
   Now we use [latex]n \ln(A) = \ln(A^n)[/latex], thinking of the division by [latex]2[/latex] as a multiplication by [latex]1/2[/latex].
   \begin{align*}
   \frac{\ln\left(1 + \frac{5}{x} \right)}{2} & = \frac{1}{2} \ln\left(1 + \frac{5}{x} \right) \\
   & = \ln \left(\left(1 + \frac{5}{x} \right)^{1/2}\right)
   \end{align*}

   (Example 2): Simplify [latex]\lim_{h \to 0} \left(1 + \frac{3h}{k}\right)^{1/h}[/latex].

   In this example, we know from our facts above that [latex]e^x = \lim_{h \to 0} \left(1 + hx \right)^{1/h}[/latex]. We also see that
   \begin{align*}
   \lim_{h \to 0} \left(1 + \frac{3h}{k}\right)^{1/h} & = \lim_{h \to 0} \left(1 + \frac{3h}{k}\right)^{1/h} \\
   & = \lim_{h \to 0} \left(1 + h \left(\frac{3}{k}\right) \right)^{1/h}
   \end{align*}
   Notice now that [latex]\frac{3}{k}[/latex] is playing the same role as [latex]x[/latex] in the [latex]e^x[/latex] equation. So this simplifies to [latex]e^{3/k}[/latex].