Question

How do I use the Limit definition of derivative on $f\left( x \right)=\tan x$.

Answer 1

The slope is equal to change in y coordinate divided by the change in x coordinate. Basically, the limit definition of derivative is the slope of the graph drawn of a function, let us say, f(x) against the infinitesimal small x changes occurring in the function. The function f(x) is plotted against the y-axis whereas the infinitesimal small changes, x are plotted against the x-axis. The limit is used, to sum up all these infinitesimal small changes with respect to the changes they are causing in the function at each instant.

Complete step by step answer:
We know that, $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ by the basic formulae of trigonometric derivatives.
Let $y=f\left( x \right)$
Differentiating both sides, we get $\dfrac{dy}{dx}=f'(x)$,
Using the limit definition of derivative,
$\Rightarrow \dfrac{dy}{dx}=f'\left( x \right)=\displaystyle \lim_{h \to 0}\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right]$
Substituting value of $f\left( x \right)=\tan x$,
$\Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\left[ \dfrac{\tan \left( x+h \right)-\tan x}{h} \right]$
Here, using the trigonometric identity: $\tan \left( a+b \right)=\left( \dfrac{\tan a+\tan b}{1-\tan a.tanb} \right)$;
$\Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\left[ \dfrac{\left( \dfrac{\tan x+\tanh }{1-\tan x.\tanh } \right)-\tan x}{h} \right]$
Taking LCM, we get:
$\Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\left[ \dfrac{\dfrac{\left( \tan x+\tanh \right)-\tan x\left( 1-\tan x.\tanh \right)}{\left( 1-\tan x.\tanh \right)}}{h} \right]$
Now, we shall subtract the similar terms in the numerator:
$\Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\left[ \dfrac{\dfrac{\tanh +{{\tan }^{2}}x.\tanh }{\left( 1-\tan x.\tanh \right)}}{h} \right]$
$\Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\left[ \dfrac{\tanh \left( 1+{{\tan }^{2}}x \right)}{h\left( 1-\tan x.\tanh \right)} \right]$
$\Rightarrow \dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\left[ \left( \dfrac{1+{{\tan }^{2}}x}{1-\tan x.\tan h} \right)\left( \dfrac{\tanh }{h} \right) \right]$
We know that $\tan x=\dfrac{\sin x}{\cos x}$, therefore, substituting this value we get:
Here, $\displaystyle \lim_{h \to 0}\left( \dfrac{\tanh }{h} \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\sinh }{\cosh .h} \right)$
On further solving, we get
$\displaystyle \lim_{h \to 0}\left( \dfrac{\tanh }{h} \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\sinh }{h} \right).\displaystyle \lim_{h \to 0}\left( \dfrac{1}{\cosh } \right)$
By the property of limits, we know that for an infinitesimally small angle $h \to 0$ , the value of limit of$\dfrac{\sinh }{h}$, that is, $\displaystyle \lim_{h \to 0}\left( \dfrac{\sinh }{h} \right)=1$
Applying this property, we get
$\displaystyle \lim_{h \to 0}{\mathop{\Rightarrow \lim }}\,\left( \dfrac{\tanh }{h} \right)=1.\displaystyle \lim_{h \to 0}\left( \dfrac{1}{\cosh } \right)$
Substituting the value of $h=0$, we get
$\displaystyle \lim_{h \to 0}{\mathop{\Rightarrow \lim }}\,\left( \dfrac{\tanh }{h} \right)=\dfrac{1}{\cos 0}$
Now, since $\cos {{0}^{\circ }}=1$, thus we substitute this value and get,
$\displaystyle \lim_{h \to 0}{\mathop{\Rightarrow \lim }}\,\left( \dfrac{\tanh }{h} \right)=1.1$
$\therefore \displaystyle \lim_{h \to 0}\left( \dfrac{\tanh }{h} \right)=1$
Therefore,
$\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\left( \dfrac{1+{{\tan }^{2}}x}{1-\tan x.\tanh } \right)$
Applying the limits,
$\therefore \dfrac{dy}{dx}=\dfrac{1+{{\tan }^{2}}x}{1-0}$
$\Rightarrow \dfrac{dy}{dx}=1+{{\tan }^{2}}x$
$\therefore \dfrac{dy}{dx}={{\sec }^{2}}x$
Therefore, by using the limit definition of derivatives, we find that the derivative of $\tan x$ is equal to ${{\sec }^{2}}x$.

Note: The limit definition of the derivative has various uses all revolving around our daily life. We can also find the derivative of $\tan x$by writing it as $\dfrac{\sin x}{\cos x}$ and then applying the quotient rule of differentiation. While solving this problem, mistakes can be made by us while dividing and multiplying multiple terms together. In order to avoid that, we must solve each step separately and carefully.