অন্তরজের লিমিট-সংজ্ঞা ব্যবহার করে ব্যবকলন

When I was young, I had to learn the fundamentals of basketball. You can have all the physical ability in the world, but you still have to know the fundamentals.

• Michael Jordan

In the group Differentiation Rules and Computations, we created nifty tools like the Product Rule and Chain Rule.

But you still have to know the fundamentals! Remember that a derivative just measures the sensitivity of a function at a point, which we can define (equivalently!) either in terms of a different quotient $\frac{f(x+h)-f(x)}{h}$ (or $\frac{f(y)-f(x)}{y-x}$ in terms of $y=x+h$) or a linear approximation $f(y) \approx f(x)+f'(x)(y-x)$.

But enough philosophy.

Practically speaking, though, is it enough to rely on the power rule, product rule, chain rule, etc.?

Not always. Here's a series of questions showing why we sometimes need to go back to the fundamentals, and use the definition of a derivative. In the following questions, let $f(x) = x(1+\lvert{x}\rvert)$.

Right. For $x>0$, we have $\lvert{x}\rvert = x$, so $f(x) = x(1+x) = x+x^2$. So by the power rule and linearity of derivatives, $f'(x) = 1+2x$ for $x>0$.

Right. For $x<0$, we have $\lvert{x}\rvert = -x$, so $f(x) = x(1-x) = x-x^2$. So by the power rule and linearity of derivatives, $f'(x) = 1-2x$ for $x<0$.

Yep. But we'll postpone the proof until later. For now, we show why the power rule, product rule, chain rule, etc. don't (at least directly) apply here:

OK, so can't use the product rule, at least not on the two factors $x$ and $1+\lvert{x}\rvert$. This is because $1+\lvert{x}\rvert$ is not differentiable at $x=0$:

We'll need to return to the basics, and study the limit $\lim_{t\to0}\frac{f(t)-f(0)}{t-0}$ (if the limit exists, then that's the derivative of $f$ at $x=0$, by definition). But to visualize some nice features of our problem, let's graph $f$, as well as some closely related functions (the $x\pm x^2$ we encountered in the first two questions above).

Hmm... $f,g,h$ all look tangent to the line $y=x$ at $(0,0)$. Equivalently, it looks like $f'(0) = g'(0) = h'(0) = 1$. Can you figure out why?

In fact, using one-sided limits, we see that $$\lim_{t\to0^+}\frac{f(t)-f(0)}{t-0} = \lim_{t\to0^+}\frac{g(t)-g(0)}{t-0} = g'(0) = 1+2\cdot0 = 1$$ and $$\lim_{t\to0^-}\frac{f(t)-f(0)}{t-0} = \lim_{t\to0^-}\frac{h(t)-h(0)}{t-0} = h'(0) = 1-2\cdot0 = 1.$$ (These are sometimes called one-sided or left and right derivatives.)

By two-sided limits in terms of one-sided limits, we obtain $$\lim_{t\to0}\frac{f(t)-f(0)}{t-0} = 1.$$ So by the limit definition, $f'(0)$ exists and equals $1$.

I hope you found this illuminating!

Usage of the limit definition to find derivatives can be a tedious exercise; it requires some complicated algebra. Generally, other techniques of differentiation are preferred, but the limit definition serves as a more rigorous proof. The limit definition is defined as

$f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$

Illustrations of how this definition is used are quite important. We will use it to prove that the derivative of $\tan(x)$ is $\sec^2(x)$.

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\tan(x + h) - \tan(x)}{h}$

Direct substitution on the right-hand side gives $\frac{0}{0}$, so another approach must be used. Use the angle addition formula for tangent on the $\tan(x + h)$ term:

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\frac{\sin(x)\cos(h) + \cos(x)\sin(h)}{\cos(x)\cos(h) - \sin(x)\sin(h)} - \tan(x)}{h}$

Simplify the complex fraction:

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\frac{\tan(x)\cos(h) + \sin(h)}{\cos(h) - \tan(x)\sin(h)} - \tan(x)}{h} \Rightarrow$

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\frac{\tan(x) + \tan(h)}{1 - \tan(x)\tan(h)} - \tan(x)}{h} \Rightarrow$

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\tan(x) + \tan(h) - \tan(x)(1 - \tan(x)\tan(h))}{h(1 - \tan(x)\tan(h))} \Rightarrow$

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\tan(x) + \tan(h) - \tan(x) + \tan^2(x)\tan(h)}{h - h\tan(x)\tan(h)}$

Some of the terms will cancel:

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\tan(h)(1 + tan^2(x))}{h - h\tan(x)\tan(h)}$

There is an identity that states that $1 + \tan^2(x) = \sec^2(x)$:

$\sec^2(x) = \lim_{h \rightarrow 0}\frac{\tan(h) \cdot \sec^2(x)}{h - h\tan(x)\tan(h)}$

Divide both sides of the equation by $\sec^2(x)$:

$1 = \lim_{h \rightarrow 0}\frac{\tan(h)}{h - h\tan(x)\tan(h)}$

At this point we must use the result that $\lim_{h\to 0} \frac{\sin (h)}{h} = 1$. (This limit almost always comes up when dealing with trig functions.) If we expand $\tan(h)$ as $\frac{\sin(h)}{\cos(h)}$ and then factor out $\frac{\sin(h)}{h}$, we get:

$1 = \lim_{h\to 0}\left(\frac{\sin (h)}{h} \cdot \frac{1}{\cos h - \tan x \sin h}\right).$

By the product rule for limits,

$1 = \left(\lim_{h\to 0} \frac{\sin(h)}{h}\right) \cdot \left(\lim_{h\to 0} \frac{1}{\cos h - \tan x \sin h}\right).$

We know that the first limit is $1$. In the second limit, we can substitute $h=0$ directly to get $1$, obtaining $1 = 1 \cdot 1$.

So the derivative statement is true. In general, these computations are heavy on algebra and sometimes trigonometry (it depends on the derivative you are calculating).

The derivative is usually defined as either $$\begin{align*} \frac d{dx}(f(x))&=\lim_{x_0\to x}\frac{f(x_0)-f(x)}{x_0-x}\\\ \text{or}\\\ \frac d{dx}(f(x))&=\lim_{\Delta x\to x}\frac{f(x+\Delta x)-f(x)}{\Delta x} \end{align*}$$ where $f(x)$ is a function with argument $x$.

Here are some problems to practice Using the Limit Definition of a Derivative. They may be simpler to do using another rule, but the definition is important, so use it! Good luck!

(Videos) Finding a Derivative Using the Definition of Derivative

by PatrickJMT

In this video by patrickJMT he shows you how to use the derivative definition to find the derivative at a point.

Recall the definition of a derivative: $$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$

The first example he looks at is finding the derivative of $f(x)=12+7x$. So we write this into our definition of a derivative. $$f'(x)=\lim_{h\rightarrow0}\frac{12+7(x+h)-(12+7x)}{h}$$ Notice if we plug in $h=0$ we get a zero in the denominator. So to find this limit we have to do some algebra until we can plug in $h=0$. So let's expand the top. $$\begin{align} &=\lim_{h\rightarrow0}\frac{12+7x+7h-12-7x}{h}\\ &=\lim_{h\rightarrow0}\frac{7h}{h}\\ &=\lim_{h\rightarrow0}7\\ &=\boxed{7} \end{align}$$ So the derivative of $f(x)=12+7x$ is $f'(x)=7$.

The next example he solves is finding the derivative of $G(t)=\dfrac{4t}{t+1}$. So we need to plug this into our definition. $$G'(t)=\lim_{h\rightarrow0}\frac{\frac{4(t+h)}{t+h+1}-\frac{4t}{t+1}}{h}$$ So we have to get like denominators in the numerator and combine these. $$\begin{align} &=\lim_{h\rightarrow0}\frac{\frac{4(t+h)\color{red}{(t+1)}}{(t+h+1)\color{red}{(t+1)}}-\frac{4t\color{red}{(t+h+1)}}{(t+1)\color{red}{(t+h+1)}}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{4t^2+4t+4th+4h}{(t+h+1)(t+1)}-\frac{4t^2+4th+4t}{(t+1)(t+h+1)}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{4t^2+4t+4th+4h-4t^2-4th-4t}{(t+h+1)(t+1)}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{4h}{(t+h+1)(t+1)}}{h}\\ &=\lim_{h\rightarrow0}\frac{4h}{(t+h+1)(t+1)}\cdot\frac1h\\ &=\lim_{h\rightarrow0}\frac{4}{(t+h+1)(t+1)} \end{align}$$ Now if we plug in $h=0$: $$\begin{align} &=\frac{4}{(t+\color{red}{(0)}+1)(t+1)}\\ &=\frac{4}{(t+1)(t+1)}\\ &=\boxed{\frac4{(t+1)^2}} \end{align}$$

So the derivative of $G(t)=\dfrac{4t}{t+1}$ is $G'(t)=\dfrac{4}{(t+1)^2}$.

In this video he goes over another example of finding the derivative using the definition.

The function he takes the derivative of is: $f(x)=x^2+x-3$. He first calculates just $f(x+h)$ in order to keep things from getting messy in the numerator of the derivative definition. $$\begin{align} f(x+h)&=(x+h)^2+(x+h)-3\\ &=x^2+2xh+h^2+x+h-3 \end{align}$$ $$\begin{align} f'(x)&=\lim_{h\rightarrow0}\frac{x^2+2xh+h^2+x+h-3-(x^2+x-3)}{h}\\ &=\lim_{h\rightarrow0}\frac{x^2+2xh+h^2+x+h-3-x^2-x+3}{h}\\ &=\lim_{h\rightarrow0}\frac{2xh+h^2+h}{h}\\ &=\lim_{h\rightarrow0}\frac{h(2x+h+1)}{h}\\ &=\lim_{h\rightarrow0}2x+h+1\\ &=2x+\color{red}{0}+1\\ &\boxed{=2x+1} \end{align}$$

(Videos) Limit Definition of Derivative for Rational Functions

by PatrickJMT

Here's a video by patrickJMT showing you how to use the limit definition of a derivative to find the derivative of a rational function.

Recall: $$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$

The problem he solves is

Find the derivative of $f(x)=\frac{3}{4x}$; what is the slope of the tangent line to $f(x)$ when $x=2$?

So we start by taking the limit definition of a derivative and applying it to our function. $$\begin{align} f'(x)&=\lim_{h\rightarrow0}\frac{\frac{3}{4(x+h)}-\frac{3}{4x}}{h} \end{align}$$ Whenever you have two fractions it's usually a good idea to combine them with like denominators. $$\begin{align} f'(x)&=\lim_{h\rightarrow0}\frac{\frac{3}{4(x+h)}\color{red}{\cdot\frac xx}-\frac{3}{4x}\color{red}{\cdot\frac{x+h}{x+h}}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{3x-3(x+h)}{4x(x+h)}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{3x-3x-3h}{4x(x+h)}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{-3h}{4x(x+h)}}{h}\\ &=\lim_{h\rightarrow0}\frac{-3h}{4x(x+h)}\cdot\frac1h\\ &=\lim_{h\rightarrow0}\frac{-3}{4x(x+h)}\\ &=\frac{-3}{4x(x+\color{red}{(0)})}\\ &\boxed{=-\frac3{4x^2}} \end{align}$$

Now we want to find the tangent line's slope at the point $x=2$. Well since the derivative is just the tangent line we can plug in $x=2$ into that. $$\begin{align} f'(2)&=-\frac{3}{4\color{red}{(2)}^2}\\ &\boxed{=-\frac3{16}} \end{align}$$

Here's another video with another example. This time the problem is:

At what point does the equation of the line tangent to the curve $y=\frac1x+2$ have a slope of $-2$?

So let's find the derivative of this function. $$\begin{align} f'(x)&=\lim_{h\rightarrow0}\frac{\frac{1}{x+h}+2-(\frac{1}{x}+2)}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{1}{x+h}\color{red}{\cdot\frac xx}-\frac{1}{x}\color{red}{\cdot\frac{x+h}{x+h}}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{x-x-h}{x(x+h)}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac{-h}{x(x+h)}}{h}\\ &=\lim_{h\rightarrow0}\frac{-h}{x(x+h)}\cdot\frac1h\\ &=\lim_{h\rightarrow0}\frac{-1}{x(x+h)}\\ &=-\frac{1}{x(x+\color{red}{(0)})}\\ &\boxed{=-\frac{1}{x^2}} \end{align}$$

Now we want to know the point where the slope is $-2$. Be careful here, this means we set our derivative equal to $-2$, not plug in $x=-2$! $$-\frac1{x^2}=-2\\ \frac1{x^2}=2\\ x^2=\frac12\\ x=\pm\sqrt{\frac12}\\ \boxed{x=\pm\frac{\sqrt2}{2}}$$

We're done now right!? Not yet! It asked for the point which means an $x$-coordinate and a $y$-coordinate. We only have the $x$ part right now. To find the $y$-coordinate we can plug in $x=\frac{\sqrt2}2$ into our original function. $$\begin{align} y&=\frac1x+2\\ y&=\frac{1}{\pm\frac{\sqrt2}2}+2\\ y&=\pm\frac{2}{\sqrt2}+2\\ y&=\pm\frac{2\sqrt2}{2}+2 \end{align}\\ \boxed{y=\pm\sqrt2+2}$$

So the points where the slope is equal to $-2$ are $$\boxed{\bigg(\frac{\sqrt2}2,\sqrt2+2\bigg)\text{ and }\bigg(-\frac{\sqrt2}2,-\sqrt2+2\bigg)}$$

(Videos) Limit Definition of Derivative for Square Roots

by PatrickJMT

Here's a video by patrickJMT where he shows you goes over an example of using the limit definition of a derivative for a square root function.

Recall: $$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$

The problem he looks at is:

Given that the derivative of $f(x)=\sqrt x$ is $f'(x)=\frac{1}{2\sqrt x}$, what is the derivative of $f(x)=\dfrac{\sqrt x}{8}$?

So let's start by putting this into our definition above. $$\begin{align} f'(x)&=\lim_{h\rightarrow0}\frac{\frac{\sqrt{x+h}}{8}-\frac{\sqrt x}{8}}{h}\\ &=\lim_{h\rightarrow0}\frac{\frac18(\sqrt{x+h}-{\sqrt x})}{h}\\ &=\frac18\lim_{h\rightarrow0}\frac{\sqrt{x+h}-{\sqrt x}}{h} \end{align}$$ You might have noticed that $\lim\limits_{h\rightarrow0}\frac{\sqrt{x+h}-{\sqrt x}}{h}$ is the derivative of $\sqrt{x}$. The problem already gave us this derivative so we can just substitute it in. $$=\frac18\frac{1}{2\sqrt x}\\ \boxed{f'(x)=\frac{1}{16\sqrt x}}$$

Notice that we ended up just multiplying the derivative of the square root of $x$, $f'(x)=\frac{1}{2\sqrt x}$, by $\frac18$.

This next video is another example with square roots, only this time he doesn't give a derivative to easily help us out.

The problem states:

Suppose $f(x)=\sqrt{2x+1}$; what is the slope of the line tangent to $f$ when $x=4$?

So we start as we always do; plugging our function into the derivative definition. $$\begin{align} f'(x)&=\lim_{h\rightarrow0}\frac{\sqrt{2(x+h)+1}-\sqrt{2x+1}}{h}\\ &=\lim_{h\rightarrow0}\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h} \end{align}$$ From here we can multiply the top and bottom by the conjugate in order to cancel out the root terms. $$\begin{align} &=\lim_{h\rightarrow0}\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}\color{red}{\cdot\frac{\sqrt{2x+2h+1}+\sqrt{2x+1}}{\sqrt{2x+2h+1}+\sqrt{2x+1}}}\\ &=\lim_{h\rightarrow0}\frac{(2x+2h+1)-(2x+1)}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}\\ &=\lim_{h\rightarrow0}\frac{2x+2h+1-2x-1}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}\\ &=\lim_{h\rightarrow0}\frac{2h}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}\\ &=\lim_{h\rightarrow0}\frac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}}\\ &=\frac{2}{\sqrt{2x+2\color{red}{(0)}+1}+\sqrt{2x+1}}\\ &=\frac{2}{2\sqrt{2x+1}}\\ &=\frac{1}{\sqrt{2x+1}} \end{align}\\ \boxed{f'(x)=\frac{1}{\sqrt{2x+1}}}$$

Now we want to find the slope of the tangent line when $x=4$. So we just plug this into our derivative function. $$f'(4)=\frac{1}{\sqrt{2\color{red}{(4)}+1}}\\ \boxed{f'(4)=\frac13}$$