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Differentiation Formnulas

Differentiation of Composite Functions

Theorem 2..3   If $y = f(u)$ and $u = g(x)$ are differentiable as a function of $u$ and $x$ 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) $

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.
$\displaystyle{1. \ y = \cos{(x^2 + x)}}$ $\displaystyle{2. \ y = \log\vert x\vert}$.

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 ,

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

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

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

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

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

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 $\displaystyle{-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}}$,

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

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} \ $

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

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

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 . We first take logarihtm to both sides. Then Next differentiate both sides to get Thus, $\displaystyle{y = x^{\alpha}}$
$\displaystyle \log{y}$$\displaystyle =$$\displaystyle \log{x^\alpha} \ (\log{p^q} = q\log p)$   $\displaystyle \log{y}$$\displaystyle =$$\displaystyle \alpha \log{x}$  

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

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

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

Example ..28   Find the derivative of $y = x^{x}$.

Take logarithm of both sides, we have SOLUTION

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

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 ..28   Differentiate $y = (\sin{x})^{x}$.

Take logarithm of both sides, we have SOLUTION

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

Differentiate both sides with respect . $x$

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

Thus

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

Differentiation of Parametric Functions

  Suppose that and are differentiable on and . Then is differentiable in and the following is holds.
Theorem ..25$x = f(t)$$y = g(t)$$I$$f^{\prime}(t) \neq 0$$y$$x$

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

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

  Given . Find
Example ..29$\displaystyle{x = a\cos{t}, y = a\sin{t}, \ a > 0}$$\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 ..29   Given . Find $\displaystyle{x = a\cos^{3}{t}, y = a\sin^{3}{t},\ a > 0}$$\displaystyle{\frac{dy}{dx}}$.

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?D

(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?D

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



   
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