Differentiation Formnulas

Differentiation of Composite Functions

Theorem 2..3 If $y = f(u)$

and

are differentiable as a function of $u$

and

respectively, then the compostite function $y = f(g(x))$

is differentiable as a function of $x$

and

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = f^{\prime}(g(x))g^{\prime}(x)$

Understanding Let $x$ be the number of plankton, $u$ be the number of minnows, and $y$ be the number of perch, then $\frac{du}{dx}$ represents the rate instantaneous change of minnows against plankton. Also, $\frac{dy}{du}$ represents the rate of instantaneous change of perch againt minnows. Thus, the rate of instantaneous change of perch against plankton can be expressed by $\frac{dy}{dx}$ .

NOTE Denote $\Delta x$ small change of $x$ . Then $u = g(x)$ changes $\Delta u = g(x + \Delta x) - g(x)$ . Also, $y = f(g(x))$ changes $\Delta y = f(g(x+\Delta x)) - f(g(x))$ . Thus, $g(x+\Delta x) = g(x) + \Delta u$ and

$\displaystyle f(g(x+\Delta x)) = f(g(x) + \Delta u).$

Therefore,

$\displaystyle \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}$	$\displaystyle =$	$\displaystyle \lim_{\Delta x \to 0}\frac{f(g(x + \Delta x)) - f(g(x))}{\Delta x}$
	$\displaystyle =$	$\displaystyle \lim_{\Delta x \to 0}\frac{f(g(x) + \Delta u) - f(g(x))}{\Delta u}\frac{\Delta u}{\Delta x}$
	$\displaystyle =$	$\displaystyle \lim_{\Delta x \to 0}\frac{f(g(x) + \Delta u) - f(g(x))}{\Delta u}\cdot \frac{g(x+\Delta x) - g(x)}{\Delta x}$

Note that $\Delta x \to 0$ implies $\Delta u \to 0$ and $y=f(u), u = g(x)$

are differentiable, we have

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx}.$

Example 2..6 Differentiate the following functions.
$\displaystyle{1. y = \cos{(x^2 + x)}}$ $\displaystyle{2. y = \log\vert x\vert}$ .

$(\cos{(x^2 + x)})' = -\sin(x^2 + x)\cdot (x^2 + x)' = -(2x+1)\sin(x^2+x)$ . SOLUTION 1. $\cos{(x^2 + x)}$ is a composite function of $u = x^2 + x$ and $y = \cos{u}$ . Thus

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = -\sin{u}(2x + 1) = -(2x+1) \sin{(x^2 + x)}\ensuremath{ \blacksquare}$

2. Suppose $x > 0$ . Then $y = \log{x}$ and $\frac{dy}{dx} = \frac{1}{x}$ . Suppose next that $x < 0$ . Then $x < 0$ and $\vert x\vert = -x$ imply $y = \log\vert x\vert = \log(-x)$ . Set $u = -x$ . Then ,

$\displaystyle \frac{dy}{dx} = \frac{\log(-x)}{dx} = \frac{\log{u}}{du}\cdot \fr... ... \frac{1}{u}\cdot (-1) = -\frac{1}{-x} = \frac{1}{x}\ensuremath{ \blacksquare}$

Derivative of Log 3. $(\log\vert x\vert)' = \frac{1}{x}$

Exercise 2..6 Let $n,m$

be integers. Differentiate the following

$\displaystyle y = x^{\frac{m}{n}}$

SOLUTION Raise both sides of the equation to the nth power.

$\displaystyle y^{n} = x^{m}$

Now differentiate both sides by $x$

. Then

$\displaystyle \frac{d(y^{n})}{dx} = \frac{d(y^{n})}{dy} \cdot \frac{dy}{dx} = ny^{n-1}\cdot \frac{dy}{dx}$

$\displaystyle \frac{d(x^{m})}{dx} = mx^{m-1}$

Thus

$\displaystyle ny^{n-1} \cdot \frac{dy}{dx} = m x^{m-1}$

From this, we obtain

$\displaystyle \frac{dy}{dx} = \frac{m}{n} \cdot \frac{x^{m-1}}{y^{n-1}} = \frac... ...^{\frac{m}{n}}}{x} = \frac{m}{n} x^{\frac{m}{n} - 1}\ensuremath{ \blacksquare}$

We raise both sides of equation to the nth power so that we can use $(x^{n})' = nx^{n-1}$ .
To find $\frac{d(y^n)}{dx}$ , we first differentiate with respect to $y$ . Then $\frac{d(y^n)}{dx} = \frac{d(y^n)}{dy}\cdot \frac{dy}{dx}$ .

Differentiation of Inverse Function

Theorem 2..4 Suppose that $y = f(x)$

is differentiable on some interval and $f^{\prime}(x) \neq 0$ . If the inverse $x = f^{-1}(y)$ of $y = f(x)$

exists, then

$\displaystyle \frac{dx}{dy} = \frac{1}{dy/dx}$

Proof Let $x_{0} = f^{-1}(y_{0})$ . Then

$\displaystyle \frac{dx}{dy} = \lim_{y \rightarrow y_{0}}\frac{f^{-1}(y) - f^{-1... ...}\frac{x - x_{0}}{f(x) - f(x_{0})} = \frac{1}{dy/dx}\ensuremath{ \blacksquare}$

Inverse In Exercise2.4, we have found the derivative of $y = \log{x}$ . But note that $y = \log{x}$ is the inverse of $y = e^{x}$ . Thus $y = \log{x}$ is the same as $x = e^y$ . By the theorem above, differentiate both sides by $y$ , we get $\frac{dx}{dy} = e^{y}$ . Therefore, $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$ .

Example 2..7 Find the derivative of the following.

$\displaystyle y = \sin^{-1}{x}$

SOLUTION Note that $y = \sin^{-1}{x} \Leftrightarrow x = \sin{y}$ for the principal value is in $\displaystyle{-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}}$ . Differentiate both sides of $x = \sin{y}$ by $y$

. Then

$\displaystyle \frac{dx}{dy} = \frac{d(\sin{y})}{dy} = \cos{y}$

Since $\cos{y} \geq 0$ for $\displaystyle{-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}}$ ,

Derivation
$\cos^2{y} + \sin^2{y} = 1$ implies that $\cos{y} = \pm \sqrt{1 - \sin^{2}{y}}$ .

$\displaystyle \cos{y} = \sqrt{1 - \sin^{2}{y}} = \sqrt{1 - x^{2}}$

Thus by the formula for differentiation of inverse function,

$\displaystyle \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{\sqrt{1 - x^{2}}} \ensuremath{ \blacksquare}$

Exercise 2..7 Find the derivative of the following.

$\displaystyle y = \tan^{-1}{x}$

SOLUTION Note that $y = \tan^{-1}{x} \Leftrightarrow x = \tan{y}$ for the principal valueof $y \in \displaystyle{-\frac{\pi}{2} < y < \frac{\pi}{2}}$ . Differentiate both sides of $x = \tan{y}$ by $y$ . Then

$\displaystyle \frac{dx}{dy} = \frac{d(\tan{y})}{dy} = \sec^{2}{y} = 1 + \tan^{2}{y} = 1+x^2.$

Derivation
Divide both side of $\cos^{2}{y} + \sin^{2}{y} = 1$ by $\cos^{2}{y}$ . Then we have $\frac{\cos^2{y}}{\cos^{2}{y}} + \frac{\sin^{2}{y}}{\cos^{2}{y}} = \frac{1}{\cos^{2}{y}}$ . Now write $\frac{1}{\cos{y}}$ as $\sec{y}$ .

How to read $\sec$ we read $\sec$ as secant.

Basic Formula $\begin{displaymath}\begin{array}{ll} 7. &(\sin^{-1}{x})' = \frac{1}{\sqrt{1-x^2}}\\ 8. &(\tan^{-1}{x})' = \frac{1}{1+x^2} \end{array}\end{displaymath}$

Then by the formula for the differentiation of inverse, we obtain

$\displaystyle \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{1+x^2}\ensuremath{ \blacksquare}$

Logarithmic Differentiation To find the derivative of $\displaystyle{y = x^{\alpha}}$ . We first take logarihtm to both sides. Then

$\displaystyle \log{y}$	$\displaystyle =$	$\displaystyle \log{x^\alpha} (\log{p^q} = q\log p)$
$\displaystyle \log{y}$	$\displaystyle =$	$\displaystyle \alpha \log{x}$

Next differentiate both sides to get

$\displaystyle \frac{1}{y}y^{\prime} = \alpha \frac{1}{x}$

Thus,

$\displaystyle y^{\prime} = y ( \alpha \frac{1}{x} ) = x^{\alpha}( \alpha \frac{1}{x} ) = \alpha x^{\alpha - 1}$

Check
$\frac{d(\log{y})}{dx} = \frac{d(\log{y})}{dy}\cdot \frac{dy}{dx} = \frac{y'}{y}$ .

NOTE The name logarithmic differentiation comes from this process. We also note that the derivative of $y = x^{\alpha}$ looks exactly the same as the derivative of $y = x^{n}$ .

Example 2..8 Find the derivative of $y = x^{x}$ .

Note that the derivative of $x^x$ can not be derived from the basic formula. Thus we use logarithmic differentiation.

Check
Note that $x\log{x}$ contains two 's. In other words, it is a product of and $\log{x}$ . Thus to find the derivative, we have to use the product rule. Then $(x\log{x})' = 1\cdot \log{x} + x\cdot \frac{1}{x} = \log{x} + 1$ .

SOLUTION Take logarithm of both sides, we have

$\displaystyle \log{y} = \log{x^{x}} = x\log{x}.$

Differentiate both sides with respect $x$

$\displaystyle \frac{y'}{y} = \log{x} + x \frac{1}{x} = 1 + \log{x}.$

Thus

$\displaystyle y' = x^{x}(1 + \log{x})\ensuremath{ \blacksquare}$

Exercise 2..8 Differentiate $y = (\sin{x})^{x}$ .

SOLUTION Take logarithm of both sides, we have

$\displaystyle \log{y} = \log{(\sin{x})^{x} = x\log{\sin{x}}}.$

Differentiate both sides with respect $x$

$\displaystyle \frac{y'}{y} = \log{\sin{x}} + x \frac{\cos{x}}{\sin{x}}$

Thus

$\displaystyle y' = (\sin{x})^{x}(\log{\sin{x}} + \frac{x\cos{x}}{\sin{x}})\ensuremath{ \blacksquare}$

Check
$\log(\sin{x})^x$ can be expressed as $x\log(\sin{x})$ . Using the formula for the differentiation of inverse, $(\log{\sin{x}})' = \frac{(\sin{x})'}{\sin{x}} = \frac{\cos{x}}{\sin{x}}$ .

Differentiation of Parametric Functions

Theorem 2..5 Suppose that $x = f(t)$

and

are differentiable on $I$

and $f^{\prime}(t) \neq 0$ . Then $y$

is differentiable in $x$

and the following is holds.

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g^{\prime}(t)}{f^{\prime}(t)}$

NOTE For a function $y = f(x)$ , the value of $y$ is determined by the value of $x$ . If you want describe the behavior of ant on a table, you want to know the position of ant. To do this, $x$ and $y$ must be expressed using the time variable $t$ . Then we say $t$ parameter. If $x$ and $y$ is given by $t$ , then by the small change of $t$ cause some change of $x$ and $y$ . The amount of change is given by $\Delta x$ and $\Delta y$ . Thus the rate of small change of $y$ with respect to small change of $x$ is given by $\frac{\Delta y}{\Delta x}$ .

Understanding $\frac{\Delta y}{\Delta x} = \frac{\frac{\Delta y}{\Delta t}}{\frac{\Delta x}{\Delta t}}$ Thus the rate of infinitesimal change is given by

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Example 2..9 Given $\displaystyle{x = a\cos{t}, y = a\sin{t}, a > 0}$ . Find $\displaystyle{\frac{dy}{dx}}$ .

SOLUTION

$\displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

implies that

$\displaystyle \frac{dy}{dx} = \frac{a\cos{t}}{-a\sin{t}} = -\frac{\cos{t}}{\sin{t}}\ensuremath{ \blacksquare}$

Exercise 2..9 Given $\displaystyle{x = a\cos^{3}{t}, y = a\sin^{3}{t}, a > 0}$ . Find $\displaystyle{\frac{dy}{dx}}$ .

$\cos^{3}{t}$ $\cos^{3}{t}$ means $(\cos{t})^3$ . Thus $(\cos^{3}{t})' = 3(\cos{t})^2\cdot(\cos{t})' = -3\sin{t}\cos^2{t}$ .

SOLUTION

$\displaystyle \frac{dy}{dx} = {\frac{dy}{dt}}/{\frac{dx}{dt}} = \frac{3a\sin^{2}{t}\cos{t}}{-3a\cos^{2}{t}\sin{t}} = -\tan{t} \ensuremath{ \blacksquare}$

Exercise A

1.: Find the derivative of the following functions using the differentiation of inverse functions
(a) $\displaystyle{y = x^{\frac{1}{n}}, x > 0}$ (b) $\displaystyle{y = \sqrt{x}, x > 0}$
2.: Find the derivative of the following functions;
(a) $\displaystyle{y = (x^{2} + 1)^{2004}}$ (b) $\displaystyle{y = (x^{2} + \frac{1}{x^{2}})^{3}}$ (c) $\displaystyle{y = [(2x+1)^{2} + (x+1)^{2}]^{3}}$
3.: Find $\frac{dy}{dx}$
(a) $\displaystyle{x = t + 1, y = t^{2}-1}$ (b) $\displaystyle{x^{2} + y^{2} = 1}$
4.: Find the derivative of the following functions:
(a) $\displaystyle{x^{2}\log{x}}$ (b) $\displaystyle{x^{3}\sin{2x}}$ (c) $\displaystyle{\sin^{-1}{(2x)}}$ (d) $\displaystyle{\sqrt{e^{x} + 1}}$ (e) $\displaystyle{(\sin(x+1))^{3}}$ (f) $\displaystyle{x\sin^{-1}(2x)}$

Exercise B

1.

Find the derivative of the following functions using the differentiation of inverse functions

(a) $\displaystyle{y = \cos^{-1}{x}}$ (b) $\displaystyle{y = \tan^{-1}{x}}$

2.

Find the derivative of the following functions using logarithmic differentiation.

(a) $\displaystyle{y = x^{2}\sqrt{\frac{x^{3} + 2x + 1}{x^{2} - 3x + 1}}}$ (b) $\displaystyle{y = x^{x}}$ (c) $\displaystyle{y = \sin({x}^{x})}$ (d) $\displaystyle{y = x^{1/x}}$

3.

Find $\frac{dy}{dx}$ .

(a) $\displaystyle{x = a\cos{t}, y = a\sin{t}, a > 0}$ (b) $\displaystyle{x = \sqrt{t} - \frac{1}{t}, y = t + \frac{1}{\sqrt{t}}}$

4.

Find the derivative of the following functions:

(a) $\displaystyle{x^{2}(1 + \sqrt{x})}$ (b) $\displaystyle{x^{3}\tan{2x}}$ (c) $\displaystyle{x\sin^{-1}{x}}$ (d) $\displaystyle{\frac{x}{x^{2}+1}}$ (e) $\displaystyle{x\sin{x}}$

(f) $\displaystyle{x\sin^{-1}x + \sqrt{1-x^{2}}}$ (g) $\displaystyle{\tan^{-1}(x^{2} + 1)}$ (h) $\displaystyle{\cos{(\sqrt{2x+1})}}$

(i) $\displaystyle{\frac{\sin{x} - x\cos{x}}{x\sin{x} + \cos{x}}}$ (j) $\displaystyle{e^{2x}\cos{x}}$ (k) $\displaystyle{\log{\vert x + \sqrt{x^{2} + A}\vert}}$ (l) $\displaystyle{y = \sin{(x^{2} + 1)}}$ (m) $\displaystyle{y = \cos{(\sqrt{x + 1})}}$ (n) $\displaystyle{y = e^{\sin{x}}}$