I personally never bothered with mathematical induction in a first calculus course (except for perhaps a very few exceptions), but I did point out how the sum rule for derivatives, $(f+g)' = f' + g',$ can be reapplied successively to get the sum rule for derivatives that textbooks actually require in exercises but almost never explicitly tell students about. For example, $$(f+g+g)' \; = \; \left( (f+g) \, + \, h \right)' \; = \; (f+g)' \, + h' \; = \; (f'+g') \, + \, h' \; = \; f' + g' + h' $$ However, I have previously collected some induction examples here, here, and here. Below I've LaTexed some of the examples that might be of use in your situation. Also, in this 16 January 2007 sci.math post is a bibliography of 72 books and papers on mathematical induction, mostly papers from elementary journals that are not reviewed by Mathematical Reviews or Zentralblatt MATH (and thus not easily found except by what I did, which was to look through every issue of several journal titles at various university libraries, which took me several years).

1. Prove that the sum rule holds for arbitrary finite sums: $$\frac{d}{dx} \left( f_1 + f_2 + f_3 + \ldots + f_n \right) \;\; = \;\; \frac{d}{dx}\left(f_{1}\right) + \frac{d}{dx}\left(f_{2}\right) + \frac{d}{dx}\left(f_{3}\right) + \ldots + \frac{d}{dx}\left(f_{n}\right)$$

2. Prove that the $n$th derivative of the product of two functions follows the pattern for expanding the $n$th power of a sum (i.e. binomial expansion for positive integer powers): $$(fg)' \; = \; f'g + fg'$$ $$(fg)'' \; = \; f''g + 2f'g' + fg''$$ $$(fg)''' \; = \; f'''g + 3f''g' + 3f'g'' + fg'''$$ Let the $0$th derivative of a function be the function itself in order to make the correspondence complete.

This also works for the $n$th derivative of a product of $m$ functions (corresponds to $n$th power of an $m$-omial), but this is probably too elaborate for your needs. Anyway, the version for $m = 2$ (what I have above) is proved almost exactly like you prove the binomial expansion using induction in a high school or college precalculus course (which I realize most people skip, including myself, except when I taught honors level precalculus in a high school).

3. Using logarithms and L'Hopital's rule, you can easily show: $$\lim_{x \rightarrow 0^{+}} x^{x} \;\; = \;\; 0 $$ $$\lim_{x \rightarrow 0^{+}} x^{x^{x}} \;\; = \;\; 1 $$ $$\lim_{x \rightarrow 0^{+}} x^{x^{x^{x}}} \;\; = \;\; 0 $$ $$\lim_{x \rightarrow 0^{+}} x^{x^{x^{x^{x}}}} \;\; = \;\; 1 $$ Using induction, you can show that this pattern continues. For an odd number of $x$'s the expression approaches $0$ as $x$ approaches $0$ from the right, and for an even number of $x$'s the expression approaches $1$ as $x$ approaches $0$ from the right.

This is something that a calculator exploration will suggest, by the way. In fact, the very first day I ever used a graphing calculator (and this also includes any computer algebra system, since I had never used one of those until several years after I first used a graphing calculator) I graphed $y = x^{x},$ then $y = x^{x^x},$ then $y = x^{x^{x^x}},$ and so on, and noticed this pattern. I even skipped ahead and tried graphing functions such as $y= x^{{x}^{{x}^{{\cdot}^{{\cdot}^{{\cdot}^{x}}}}}}$ $(8\; x\text{'s})$ and $y = x^{{x}^{{x}^{{\cdot}^{{\cdot}^{{\cdot}^{x}}}}}}$ $(15\; x\text{'s}).$ I did this on a HP-28S which I had purchased that day, sometime back in Fall 1988 or Winter 1988-89 (I don't remember more precisely now).

4. Let $x$ be a real number. Define $x\,\#\,0 = 1$ and, for positive integers $n,$ define $x\,\#\,n \; = \; x(x-1)(x-2)\cdots(x-n+1).$ Then using induction one can prove that the forward difference operator applied to $x\,\#\,n\;$ is $\;n\cdot[x\,\#\,(n-1)].$ Note: The "forward difference operator" applied to the function $f(n),\;$ where $f:{\mathbb N} \rightarrow {\mathbb R},\;$ gives the result $f(n+1) - f(n).$

5. Find an error in the following "proof":

We prove by induction that $\frac{d}{dx}(x^n) = 0$ for each $n = 0,$ $1,$ $2,\, \ldots $ Note that $\frac{d}{dx}(x^0)=0,$ since the derivative of a constant is zero. Now assume $\frac{d}{dx}(x^k)=0$ for each $k \leq n.$ (We'll be using what's sometimes called "strong induction" or "course of values induction".) Then $\frac{d}{dx}(x^{n+1})=0,$ as shown below: $$ \frac{d}{dx}\left( x^{n+1} \right) \;\; = \;\; \frac{d}{dx}\left( x^{n} \cdot x \right) \;\; = \;\; \frac{d}{dx}\left(x^{n}\right) \cdot x \; + \; x^{n} \cdot \frac{d}{dx} (x) \;\; = \;\; 0 \cdot x \; + \; x^{n} \cdot 0 \;\; = \;\; 0 $$

6. Let $(d/dx)^n$ be $n$-many differentiations with respect to $x.$ $$\frac{d^n}{dx^n}\left( \ln x \right) \;\; = \;\; (-1)^{n-1}(n-1)! \cdot x^{-n} $$ $$\frac{d^n}{dx^n}\left( a^{b^x} \right) \;\; = \;\; b^{n} (\ln a)^{n} a^{bx} $$ $$\frac{d^n}{dx^n}\left( \sin (ax+b) \right) \;\; = \;\; a^{n} \sin \left( ax+b + \frac{n\pi}{2} \right) $$ $$\frac{d^n}{dx^n}\left( \cos (ax+b) \right) \;\; = \;\; a^{n} \cos \left( ax+b + \frac{n\pi}{2} \right) $$ $$\frac{d^n}{dx^n}\left( e^{x}\cos x \right) \;\; = \;\; \left(\sqrt{2}\right)^{n}e^{x} \cos \left(x + \frac{n\pi}{4} \right) $$ $$\frac{d^n}{dx^n}\left( \sqrt{x+1} \right) \;\; = \;\; (-1)^{n-1} \cdot \frac{(1)(3)(5)\cdots (2n-3)}{2^{n}(x+1)^{n - \frac{1}{2}}} $$ $$\frac{d^n}{dx^n}\left( x^2 - 1 \right)^{-1} \;\; = \;\; (-1)^{n} \cdot \frac{n!}{2} \left[ (x-1)^{-n-1} \; - \; (x+1)^{-n-1} \right] $$ $$\frac{d^n}{dx^n}\left( \frac{3x-4}{2x^2 + 3x -2 } \right) \;\; = \;\; (-1)^{n} \cdot n! \left[ \frac{2^{n+1}}{(2x-3)^{n+1}} \; - \; \frac{3^{n+1}}{(3x+2)^{n+1}} \right] $$

For the last two, develop a formula for the $n$th derivative of $(ax + b)^{-1}$ and then use partial fractions (and additivity of the $n$th derivative operation).

From what I've read in old texts (> 100 years old), these kinds of results were often useful in calculating Taylor coefficients. In particular, the formulas often allowed one to come up with recursion formulas for the $n$th derivative evaluated at zero, which allowed for more rapid calculation of the coefficients than multiplying known power series expansions together.