Use the quotient rule and the derivatives of $\sin x$ and $\cos x$ to show $\displaystyle{\frac{d}{dx} \tan x = \sec^2 x}$.

$\sec^2 x$

Find $\displaystyle{\frac{d}{dx}(\sec x \cot x - \csc x)}$

$0$

Find $\displaystyle{\frac{d}{dx}(x e^x \sin x)}$

$e^x(x \cos x + x \sin x + \sin x)$

Find the following limits.

$\displaystyle{\lim_{\theta \rightarrow 0} \frac{1-\cos\theta}{\theta}}$

$\displaystyle{\lim_{t \rightarrow 0} \frac{\sin 3t}{2t}}$

$\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos x}{x^2}}$

$0$

$\displaystyle{\frac{1}{2}}$

$\displaystyle{\frac{1}{2}}$

Prove the derivative rules $\displaystyle{\frac{d}{dx} \sin x = \cos x}$ and $\displaystyle{\frac{d}{dx} \cos x = -\sin x}$.

Use the identities $\sin(x+h) = (\sin x)(\cos h) + (\sin h)(\cos x)$ and $\cos(x+h) = (\cos x)(\cos h) - (\sin x)(\sin h)$ in the limit definitions for the derivatives of the sine and cosine functions, respectively.

Then, in both, isolate a $\displaystyle{\frac{\sin x}{x}}$ inside the body of the limit.

Remember that expressions entirely in terms of $x$, in a limit where $h$ is approaching some value, can be thought of as constant and treated accordingly.

Compute the 6

^{th}derivative of $f(x) = e^x \sin x$. Based on the pattern seen in the derivatives, what should $f^{(15)}(x)$ be?The sixth derivative is $-8e^x \cos x$, while $f^{(15)} = -128(\cos x - \sin x)$.

Use induction to show that $\displaystyle{\frac{d^n}{dx^n} (x^n) = n!}$, where $n!$, called the factorial of $n$, is the product of all the numbers from $1$ to $n$. Note that $(k+1)! = (k+1) \cdot k!$ for any integer $k \ge 0$.

Assuming the inductive hypothesis that $\displaystyle{\frac{d^k}{dx^k} (x^k) = k!}$, consider the consequences of

$$\frac{d^{k+1}}{dx^{k+1}} \left(x^{k+1}\right) = \frac{d^k}{dx^k} \left( \frac{d}{dx} \left(x^{k+1}\right) \right)$$