A review of differentiation

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We review differentiation and integration.

Review of derivative rules

One of the fundamental objects of differential calculus is the derivative. As a reminder, here are some important results:

Basic Derivative Rules Let $n\ne 0$ , $k$ be a constant and $a>0$ .

Powers of $x$ :

$\ddx k =0$
$\ddx x^n = \answer [given]{n x^{n-1}}$

Exponentials:

$\ddx e^x = \answer [given]{e^x}$
$\ddx a^x = a^x\ln (a)$

Logarithms:

$\ddx \ln (x) = \frac {1}{x}$
$\ddx \log _a (x) = \frac {1}{\ln a} \frac {1}{x}$

Trigonometric Functions:

$\ddx \sin (x) = \answer [given]{\cos (x)}$
$\ddx \cos (x) = \answer [given]{-\sin (x)}$
$\ddx \tan (x) = \sec ^2(x)$
$\ddx \sec (x) = \answer [given]{\sec (x)\tan (x)}$
$\ddx \csc (x) = -\csc (x)\cot (x)$
$\ddx \cot (x) = \answer [given]{-\csc ^2(x)}$

Inverse Trigonometric Functions:

$\ddx \arcsin (x) = \frac {1}{\sqrt {1-x^2}}$
$\ddx \arccos (x) = \frac {-1}{\sqrt {1-x^2}}$
$\ddx \arctan (x) =\frac {1}{1+x^2}$

Using the above rules only, what is the derivative of $\frac {1}{\sqrt [4]{x^3}}$ with respect to $x$ ? $\ddx \left [\frac {1}{\sqrt [4]{x^3}}\right ] = \answer [given]{- \frac {3}{4} x^{-7/4}}$

Once the derivatives of basic functions have been established, we can use them to build the derivatives of more complicated functions:

Derivatives of combinations of basic functions Let $f(x)$ and $g(x)$ be differentiable functions.

Addition: $\ddx \left ( f(x) + g(x) \right ) = f'(x) + g'(x)$
Scalar Multiplication: $\ddx \left ( af(x) \right ) = a f'(x)$
Product Rule: $\ddx \left ( f(x) \cdot g(x) \right ) = f(x)g'(x) + f'(x)g(x)$
Quotient Rule: $\ddx \frac {f(x)}{g(x)} = \frac {f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$
Chain Rule: $\ddx f(g(x)) = f'(g(x)) \cdot g'(x)$

What is the derivative of $4x^2 +\frac {2}{3\sqrt {x}}$ with respect to $x$ ? $\ddx \left (4x^2 +\frac {2}{3\sqrt {x}} \right ) = \answer [given]{8x -\frac {1}{3} x^{-3/2}}$

What is the derivative of $x^2 \sin (x)$ with respect to $x$ ? $\ddx x^2 \sin (x) = \answer [given]{2x \sin (x) +x^2 \cos (x)}$

Derivatives and Tangent Lines

While there are many useful applications of derivatives, one of the most important ones is the relationship between derivatives and tangent lines.

Suppose that $f(x)$ is a differentiable function on an interval $I$ . If we pick a value $a$ in the interval, and “zoom in” on the graph around $x=a$ , notice what happens:

As evidenced by the image, when the function $f(x)$ is differentiable at a given $x$ -value, the graph of $f(x)$ becomes closer to a line as we “zoom in,” and we call this line the tangent line at $x=a$ .

To find the equation of this line, we need a point of the line and the slope of the line. The slope of the line is $m_{tan} =f'(a)$ and the point on the line is $(a,f(a))$ .

This result is also a direct consequence of the limit definition of the derivative and some algebra. If $f'(a)$ exists, then:

$f'(a) = \lim _{x \to a} \frac {f(x)-f(a)}{x-a}$ In order to write everything as a single limit, note we can write $f'(a)$ as:

$\begin{align*} f'(a) &= \lim_{x \to a} f'(a)\\ &= \lim_{x \to a} \left(f'(a) \cdot \frac{x-a}{x-a}\right)\\ &= \lim_{x \to a} \frac{f'(a)(x-a)}{x-a} \end{align*}$

We now return to the limit definition of $f'(a)$ and write:

$\begin{align*} f'(a) &= \lim_{x \to a} \frac{f(x)-f(a)}{x-a}\\ \lim_{x \to a} \frac{f'(a)(x-a)}{x-a} &= \lim_{x \to a} \frac{f(x)-f(a)}{x-a}\\ 0 &= \lim_{x \to a} \frac{f(x)-f(a)}{x-a} - \lim_{x \to a} \frac{f'(a)(x-a)}{x-a}\\ 0 &= \lim_{x \to a} \left(\frac{f(x)-f(a)}{x-a} - \frac{f'(a)(x-a)}{x-a} \right)\\ 0 &= \lim_{x \to a} \frac{f(x)-f'(a)(x-a) - f(a)}{x-a}\\ \end{align*}$ Denoting $f'(a)(x-a) + f(a)$ by $l_a(x)$ , we note that $l_a(x)= f'(a)(x-a) + f(a)$ is the equation of a line, and we can write the last equation above as:

$\lim _{x \to a} \frac {f(x)-l_a(x)}{x-a} = 0.$ Since this limit is $0$ , we can interpret it geometrically by saying that as the distance between $x$ and $a$ becomes smaller, the distance between the function $f(x)$ and the line $l_a(x)$ becomes even smaller. That is, as we “zoom in” on the graph of a function at a point where it is differentiable, the graph becomes closer and closer to a line and we call this line the tangent line.

Many students mistakenly think that the tangent line can only touch the graph of the function at the point of tangency, but this is not true. For instance, if $f(x) = 2x+1$ , then $f'(x) = \answer [given]{2}$ and $f(1) = \answer [given]{3}$ , so the equation of the tangent line at $x=1$ is: $\begin{align*} y-y(1) &= m_{tan}(x-x(1)) \\ y - \answer[given]{3} &= \answer[given]{2}\left(x-\answer[given]{1}\right) \\ y &= \answer[given]{2x+1} \end{align*}$

This is exactly the original function. Thus, in any interval around $x=1$ , the graph of the tangent line and the function intersect onceonly finitely many timesat every point .

As such, it is important to realize that a good intuitive way to think about tangent lines is that a function has a tangent line at a point $x=a$ if the graph of the function becomes less distinguishable from its tangent line as we continue to “zoom in” on the graph of the function at $x=a$ .