Continuity and Differentiability

One of the chief features in the behaviour of functions is the property known as continuity. For instance, the continuous expansion of a rod on heating, of the continuous growth of an organism, of a continuous flow, or a continuous variation of atmospheric temperature, etc. The idea of continuity of a function stems from the geometric notion of 'no breaks in a graph'. The limit used to define the slope of a tangent line or the instantaneous velocity of a freely falling body is also used to define one of the two fundamental operations of calculus – differentiation.

This article is about the concept of class 12 maths continuity and differentiability. Continuity and differentiability chapter is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE, BCECE, and more.

Continuity and Differentiability

Continuity and differentiability is one of the fundamental concepts of calculus. They help analyze changes, optimize processes, and predict trends in fields like engineering, physics, and economics. Continuity and differentiability allow us to study functions near critical points, even if the function is not defined at those points.

Continuity

Suppose $f$ is a real function on a subset of the real numbers and let $c$ be a point in the domain of $f$. Then $f$ is continuous at $c$ if

$\underset{x\to c}{lim}f(x)=f(c)$

More elaborately, if the left-hand limit, right-hand limit, and the value of the function at $x=c$ exist and are equal to each other, then $f$ is said to be continuous at $x=c$.

If the right-hand and left-hand limits at $x=c$ coincide, then we say that the common value is the limit of the function at $\mathrm{x}=\mathrm{c}$. Hence we may also rephrase the definition of continuity as follows:

A function is continuous at $\mathrm{x}=\mathrm{c}$ if the function is defined at $x=c$ and if the value of the function at $x=c$ equals the limit of the function at $x=c$.

If $f$ is not continuous at $c$, we say $f$ is discontinuous at $c$, and $c$ is called a point of discontinuity of $f$.

If the function is continuous, Its graph does not break but for discontinuous functions, there is a break in the graph. A real function is continuous at a fixed point if we can draw the graph of the function around that point without lifting the pen from the plane of the paper. In case one has to lift the pen at a point, the graph of the function is said to have a break or discontinuity at that point, say $x=a$.

Continuity can be defined in two ways: Continuity at a point and Continuity over an interval.

Continuity at a point

Let us see different types of conditions to see continuity at point $x=a$

We see that the graph of $f(x)$ has a hole at $x=a$, which means that $f(a)$ is undefined. At the very least, for $f(x)$ to be continuous at $x=a$, we need the following conditions:

(i) $f(a)$ is defined

Next, for the graph given below, although $f(a)$ is defined, the function has a gap at $x=a$. In this graph, the gap exists because lim $\underset{x\to a}{lim}f(x)$ does not exist. We must add another condition for continuity at $x=a$, which is

(ii) $\underset{x\to a}{lim}f(x)$ exists

The above two conditions by themselves do not guarantee continuity at a point. The function in the figure given below satisfies both of our first two conditions but is still not continuous at $a$. We must add a third condition to our list:

(iii) $\underset{x\to a}{lim}f(x)=f(a)$

So, a function $f(x)$ is continuous at a point $x=a$ if and only if the following three conditions are satisfied:

i) $f(a)$ is defined

ii) $\underset{x\to a}{lim}f(x)$ exists

iii) $\underset{x\to a}{lim}f(x)=f(a)$ or

$\begin{array}{rl}& \underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)=f(a)\\ & \text{ i.e. }\text{ L.H.L. }=\text{ R.H.L. }=\text{ value of the function at }x=a\end{array}$

A function is discontinuous at a point $a$ if it fails to be continuous at $a$.

Continuity over an Interval

Over an open interval $(a,b)$: A function $f(x)$ is continuous over an open interval $(a,b)$ if $f(x)$ is continuous at every point in the interval.

For any $c\in (a,b),f(x)$ is continuous if $\underset{x\to c^{-}}{lim}f(x)=\underset{x\to c^{+}}{lim}f(x)=f(c)$

Over a closed interval $[a,b]$: A function $f(x)$ is continuous over a closed interval of the form $[a,b]$ if
- it is continuous at every point in $(a,b)$ and
- is right-continuous at $x=a$ and
- is left-continuous at $x=b$.
i.e.At $\mathrm{x}=\mathrm{a}$, we need to check $f(a)=\underset{x\to a^{+}}{lim}f(x)(=\underset{h\to 0^{+}}{lim}f(a+h)=$ R.H.L. $)$.

L.H.L. should not be evaluated to check continuity of the first element of the interval, $x=a$

Similarly, at $\mathrm{x}=\mathrm{b}$, we need to check $f(b)=\underset{x\to b^{-}}{lim}f(x)(=\underset{h\to 0^{+}}{lim}f(b-h)=$ L.H.L. $)$.

R.H.L. should not be evaluated to check continuity of the last element of the interval $x=b$

Consider one example,

$f(x)=[x]$, prove that this function is not continuous in $[2,3]$,

Solution:

Condition 1: For continuity in $(2,3)$
At any point $x=c$ lying in $(2,3)$,
$f(c)=[c]=2($ as $c$ lies in $(2,3))$
LHL at $\mathrm{x}=\mathrm{c}:x\to {\mathrm{c}}^{-}[x]=2$ (as in close left neighbourhood of $\mathrm{x}=\mathrm{c}$, the function equals 2)

RHL at $\mathrm{x}=\mathrm{c}:x\to {\mathrm{c}}^{+}[x]=2$ (as in close right neighbourhood of $\mathrm{x}=\mathrm{c}$, the function equals 2)

So function is continuous for any c lying in $(2,3)$. Hence the function is continuous in $(2,3)$

Condition 2: Right continuity at $x=2$

$\begin{array}{rl}& f(2)=2\\ & \underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}[x]=\underset{h\to 0^{+}}{lim}[2+h]=2\end{array}$

So $f(x)$ is left continuous at $x=2$

Condition 3: Left continuity at $x=3$
$f(3)=3$ and

$\underset{x\to 3^{-}}{lim}[x]=\underset{h\to 0^{+}}{lim}[3-h]=2$

(as in left neghbourhood of $3,f(x)=2$ )
So $f(3)$ does not equal LHL at $x=3$
hence $f(x)$ is not left continuous at $x=3$

So the third condition is not satisfied and hence $f(x)$ is not continuous in $[2,3]$

Properties of Continuous function

1. If $f,g$ are two continuous functions at a point a of their common domain D. Then $f\pm g$ fg are continuous at a and if $g(a)\ne 0$ then $\underset{―}{f}$
$g$ is also continuous at $\mathrm{x}=\mathrm{a}$.
Suppose $f$ and g be two real functions continuous at a real number c .
Then

(1) $f+g$ is continuous at $x=c$.
(2) $f-g$ is continuous at $x=c$.
(3) $f\cdot g$ is continuous at $x=c$.
$\left(\frac{f}{g}\right)$
(4) $\frac{g}{g}$ is continuous at $\mathrm{x}=\mathrm{c}$, (provided $\mathrm{g}(\mathrm{c})\ne 0$ ).
The sum, difference, product, and quotient of two continuous functions are always a continuous function. However $h(x)=\frac{f(x)}{g(x)}$ is continous function at $\mathrm{x}=\mathrm{a}$ only if $g(a)\ne 0$

2. If f is continuous at a and $f(a)\ne 0$ then there exists an open interval $(a-\delta ,a+\delta$ ) such that for all $xϵ(a-\delta ,a+\delta )f(x)$ has the same sign as $f(a)$

3. If a function $f$ is continuous on a closed interval $[a,b]$, then it is bounded on ( $\mathrm{a},\mathrm{b}$ ) and there exists real numbers k and K such that $k\le f(x)\le K$ for all $x\in [a,b]$

Differentiability

The instantaneous rate of change of a function with respect to the independent variable is called derivative. Let $f(x)$ be a given function of one variable and $\mathrm{\Delta }x$ denotes a number ( positive or negative) to be added to the number $x$.

Let $\mathrm{\Delta }f$ denotes the corresponding change of f , then $\mathrm{\Delta }f=f(x+\mathrm{\Delta }x)-f(x)$

If $\frac{\mathrm{\Delta }f}{\mathrm{\Delta }x}$ approaches a limit as $\mathrm{\Delta }x$ approaches to zero, this limit is the derivative of $f$ at the point $x$. The derivative of a function $f$ is denoted by symbols such as $f^{\prime }(x),\frac{\mathrm{d}f}{\mathrm{d}x},\frac{\mathrm{d}}{\mathrm{d}x}f(x)$ or $\frac{\mathrm{d}f(x)}{\mathrm{d}x}$.

The derivative of a function $f(x)$ at point $a$ is defined by$f^{\prime }(x)\text{ at }\mathrm{x}=\text{ a i.e, }{f}^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}\text{ or }\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$

If $\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$ does not exists, we say that the function $\mathrm{f}(\mathrm{x})$ is not differentiable at $x=a$.

Or, we can say that a function $f(x)$ is differentiable at a point ' $a$ ' in its domain if limit of the function $f^{\prime }(x)$ exists at $x=a$.

i.e. Right hand limit = RHD= $R{f}^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$
Left hand limit = LHD= $L{f}^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a-h)-f(a)}{-h}$

(Both the left-hand derivative and the right-hand derivative are finite and equal.)

$R{f}^{\prime }(a)=L{f}^{\prime }(a)$ is the condition for differentiability at $x=a$

$\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}=\underset{h\to 0}{lim}\frac{f(a-h)-f(a)}{-h}$

Examining the Differentiability

1. Using Differentiation (only for continuous functions)

Some functions are defined piecewise, in such cases first we need to check if the function is continuous at the split point, and if it is continuous we need to differentiate each branch function and compare left-hand and right-hand derivative at the split point.

$f(x)=\{\begin{array}{ll}{g}_1(x),& x<a\\ g_2(x),& x\ge a\end{array}$
First check if $f(x)$ is continuous at $x=a$. If it is not continuous, then it cannot be differentiable. If it is continuous, then to check differentiability, find

$f^{\prime }(x)=\{\begin{array}{ll}{(g_1(x))}^{\prime },& x<a\\ {(g_2(x))}^{\prime },& x>a\end{array}$
Differentiability can be checked at $\mathrm{x}=\mathrm{a}$ by comparing

$\underset{x\to a^{-}}{lim}{(g_1(x))}^{\prime }\text{ and }\underset{x\to a^{+}}{lim}{(g_2(x))}^{\prime }$

2. Differentiability using Graphs

A function $f(x)$ is not differentiable at $x=a$ if
1. The function is discontinuous at $x=a$
2. The graph of a function has a sharp turn at $x=a$
3. A function has a vertical tangent at $x=a$

Illustration 1

Check the differentiability of the following function.
1. $f(x)=\sin |x|$

Method 1

Using graphical transformation, we can draw its graph

Using the graph we can tell that at $x=0$, the graph has a sharp turn, so it is not differentiable at $x=0$.

Method 2

As $LHL=RHL=f(0)=0$, so the function is continuous at $x=0$
So we can use differentiation to check differentiability

$\begin{array}{rl}& \mathrm{f}(\mathrm{x})=\{\begin{array}{cc}-\sin x,& x<0\\ \sin x,& x\ge 0\end{array}\\ & \therefore {\mathrm{f}}^{\prime }(\mathrm{x})=\{\begin{array}{cc}-\cos x,& x<0\\ \cos x,& x>0\end{array}\\ & \therefore \text{ LHD }={\mathrm{f}}^{\prime }\left(0^{-}\right)=-1\text{ and }\mathrm{RHD}={\mathrm{f}}^{\prime }\left(0^{+}\right)=1\end{array}$
As these are not equal, so, $f(x)=\sin |x|$ is not differentiable at $x=0$

Illustration 2

$f(x)=\Vert \log \mid x\Vert ,x\text{ not equal to }0$
Plot the graph of | log $|\mathrm{x}|$ | using graphical transformation

We can see that graph has a sharp turn at +1 and -1 so the function is not differentiable at these points.

Properties of Differentiability

1. A function $f(x)$ is differentiable in an open in interval ( $a,b$ ) if it is differentiable at every point on the open interval $(a,b)$.

2. If $f(x)$ is differentiable at every point on the open interval (a,b). And, It is differentiable from the right at " $a$ " and the left at " $b$ ".(In other words, $\underset{x\to a^{+}}{lim}\frac{f(x)-f(a)}{x-a}$ and $\underset{x\to b^{-}}{lim}\frac{f(x)-f(b)}{x-b}$ both exists), then $f(x)$ is said to be differentiable in $[a,b]$

3. If a function $f(x)$ is differentiable at every point in an interval, then it must be continuous in that interval. But the converse may or may not be true.

Continuity and Differentiability Formulas

Continuity and differentiability class 12 formulas include algebraic properties of continuity and discontinuity, algebraic properties of differentiability, differentiation of implicit functions, and differentiation of functions in parametric form.

Algebra of Continuity and Discontinuity

The algebraic properties of continuity and discontinuity are,

1. If $f(x)$ and $g(x)$ are continuous functions in the given interval, then the following functions are continuous at $\mathrm{x}=\mathrm{a}$.
   (i) $f(x)\pm g(x)$
   (ii) $\mathrm{f}(\mathrm{x})\cdot \mathrm{g}(\mathrm{x})$
   (iii) $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}$, provided $\mathrm{g}(\mathrm{a})\ne 0$

2. If $f(x)$ is continuous and $g(x)$ is discontinuous, then $f(x)\pm g(x)$ is a discontinuous function.

   Let $f(x)=x$, which is continuous at $x=0$ and $g(x)=[x]$ (greatest integer function) which is discontinuous at $x=0$, are to function.

   Now, $f(x)-g(x)=x-[x]=\{x\}$ (fractional part of $x$ )
   discontinuous at $\mathrm{x}=0$

3. If $f(x)$ is continuous and $g(x)$ is discontinuous at $x=a$ then the product of the functions, $h(x)=f(x)g(x)$ is may or may not be continuous at $x=a$.

   For example,
   Consider the functions, $f(x)=x^3$. And $g(x)=\mathrm{sgn}(x)$.
   $f(x)$ is continuous at $x=0$ and $g(x)$ is discontinuous at $x=0$
   Now,

   $h(x)=f(x)\cdot g(x)=\{\begin{array}{cl}{x}^3,& x>0\\ 0,& x=0\\ -x^3,& x<0\end{array}$

   $h(x)$ is continuous at $x=0$
   Take another example, consider $f(x)=x$ and $g(x)=1/|x|$
   $f(x)$ is continuous at $x=0$ and $g(x)$ is discontinuous at $x=0$
   Now,

   $h(x)=f(x)\cdot g(x)=x\cdot \frac{1}{|x|}=\mathrm{sgn}(\mathrm{x})$
   And we know that signum function is discontinuous at $x=0$.

4. If $f(x)$ and $g(x)$, both are discontinuous at $x=a$ then the the function obtained by algebraic operation of $f(x)$ and $g(x)$ may or may not be continuous at $x=a$.

Rules of Differentiation

Let $f(x)$ and $g(x)$ be differentiable functions and $k$ be a constant. Then each of the following rules of differentiation holds.

Sum Rule

The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g$.

$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$
In general,

$\frac{d}{dx}(f(x)+g(x)+h(x)+\dots \dots )=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))+\frac{d}{dx}(h(x))+\dots \dots$

Difference Rule

The derivative of the difference of a function $f$ and $a$ function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$.

$\begin{array}{rl}& \frac{d}{dx}(f(x)-g(x))=\frac{d}{dx}(f(x))-\frac{d}{dx}(g(x))\\ & \frac{d}{dx}(f(x)-g(x)-h(x)-\dots \dots )=\frac{d}{dx}(f(x))-\frac{d}{dx}(g(x))-\frac{d}{dx}(h(x))-\dots \dots \end{array}$

Constant Multiple Rule

The derivative of a constant $k$ multiplied by a function $f$ is the same as the constant multiplied by the derivative of $f$

$\frac{d}{dx}(kf(x))=k\frac{d}{dx}(f(x))$

Product rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then,

$\frac{d}{dx}(f(x)g(x))=g(x)\cdot \frac{d}{dx}(f(x))+f(x)\cdot \frac{d}{dx}(g(x))$

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Extending the Product Rule

If three functions are involved, i.e let $k(x)=f(x)\cdot g(x)\cdot h(x)$
Let us have a function $\mathrm{k}(\mathrm{x})$ as the product of the function $\mathrm{f}(\mathrm{x}),\mathrm{g}(\mathrm{x})$ and $\mathrm{h}(\mathrm{x})$. That is, $k(x)=(f(x)\cdot g(x))\cdot h(x)$. Thus,

$k^{\prime }(x)=\frac{d}{dx}(f(x)g(x))\cdot h(x)+\frac{d}{dx}(h(x))\cdot (f(x)g(x))$

[By applying the product rule to the product of $f(x)g(x)$ and $h(x)$.]

$\begin{array}{rl}& =(f^{\prime }(x)g(x)+g^{\prime }(x)f(x))h(x)+h^{\prime }(x)f(x)g(x)\\ & =f^{\prime }(x)g(x)h(x)+f(x)g^{\prime }(x)h(x)+f(x)g(x)h^{\prime }(x)\end{array}$

Quotient Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x)\cdot \frac{d}{dx}(f(x))-f(x)\cdot \frac{d}{dx}(g(x))}{(g(x){)}^2}$
OR
if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime }(x)=\frac{f^{\prime }(x)g(x)-g^{\prime }(x)f(x)}{(g(x){)}^2}$
As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives.

Chain Rule

If $u(x)$ and $v(x)$ are differentiable funcitons, then $uov(x)$ or $u[v(x){]}_{\text{is also differentiable. }}$
If $y=uov(x)=u[v(x)]$, then

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u\{v(x)\}}{\mathrm{d}\{v(x)\}}\times \frac{\mathrm{d}}{\mathrm{d}x}v(x)$

is known as the chain rule. Or,

$\text{ If }y=f(u)\text{ and }u=g(x)\text{, then }\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot \frac{\mathrm{d}u}{\mathrm{d}x}$
The chain rule can be extended as follows
If $y=[\mathrm{uovow}(x)]=u[v\{w(x)\}]$, then

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}[u[v\{w(x)\}]}{\mathrm{d}v\{w(x)\}}\times \frac{\mathrm{d}[v\{w(x)\}]}{\mathrm{d}w(x)}\times \frac{\mathrm{d}[w(x)]}{\mathrm{d}x}$

Algebra of Differentiability

The algebraic properties of differentiablity is

1. If $f(x)$ and $g(x)$ are both differentiable functions at $x=a$, then the following functions are also differentiable at $\mathrm{x}=\mathrm{a}$.
(i) $\mathrm{f}(\mathrm{x})\pm \mathrm{g}(\mathrm{x})$
(ii) $\mathrm{f}(\mathrm{x})\cdot \mathrm{g}(\mathrm{x})$
(iii) $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}$, provided $\mathrm{g}(\mathrm{a})\ne 0$

2. If $f(x)$ is differentiable at $x=a$ and $g(x)$ is not differentiable at $x=a$, then $f(x)\pm g(x)$ will not be differentiable at $x=a$.

For example, $\cos (x)+|x|$ is not differentiable at $x=0$, as $\cos (x)$ is differentiable at $x=0$, but $|x|$ is not differentiable at $x=0$.

In other cases, $f(x)\cdot g(x)$ and $f(x)/g(x)$ may or may not be differentiable at $x$ = a, and hence should be checked using LHD-RHD, continuity or graph.

For example, if $f(x)=0$ (differentiable at $x=0$ ) and $g(x)=|x|$ (nondifferentiable at $x=0$ ). Their product is $f(x)\cdot g(x)=0$ which is differentiable. But if $f(x)=2,g(x)=|x|$, then $f(x)\cdot g(x)=2|x|$ is non-differentiable at $x=0$.

3. If $f(x)$ and $g(x)$ both are nondifferentiable functions at $x=a$, then the function obtained by the algebraic operation of $f(x)$ and $g(x)$ may or may not be differentiable at $x=a$. Hence they should be checked.

For example, Let $f(x)=|x|$, not differentiable at $x=0$, and $g(x)=-|x|$ which is also not differentiable at $x=0$. Their sum $=0$ is differentiable and the difference $=2|x|$ is not differentiable. So there is no definite rule.

4. Differentiation of a continuous function may or may not be continuous.

Differentiation of Implicit Functions

An implicit function is a function that includes both dependent and independent variables such as $F(x,y)=0$. For example, the equation of a circle $x^2+y^2=r^2$ is an implicit function because $y$ is not explicitly expressed as a function of $x$. Implicit differentiation means differentiation on both sides of the equation concerning the independent variable with the help of the Chain rule.

To find $\frac{dy}{dx}$ in such a case, we differentiate both sides of the given relation concerning $x$ keeping in mind that the derivative of $\mathrm{\Phi }(\mathrm{y})$ concerning $x$ is $\frac{d\varphi }{dy}\times \frac{dy}{dx}$

For example

$\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx},\frac{d}{dx}\left(y^2\right)=2y\frac{dy}{dx}$

It should be noted that $\frac{d}{dy}(\sin y)=\cos y$ but $\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}$

Differentiation of Functions in Parametric Form

Parametric differentiation is the process of finding the derivative of the equation in which the dependent variable $y$ and independent variable $x$ are equated to another variable $t$. To find the derivative of $y$ for $x$ we use the chain rule.

Sometimes, $x$ and $y$ are given as functions of a single variable, i.e., $x=g(t)$ and $y=f(t)$ are two functions and $t$ is a variable. In such cases, $x$ and $y$ are called parametric functions or parametric equations and $t$ is called the parameter.

To find $\frac{dy}{dx}$ in such cases, first find the relationship between $x$ and $y$ by eliminating the parameter $t$ and then differentiate concerning $t$.

We have $x=f(t)$ and $y=f(t)$, in this case, we will differentiate both functions separately. We will first differentiate $x$ and $y$ concerning '$t$ separately. On differentiating $x$ for ' $t$ ' we get $\frac{dx}{dt}$ and on differentiating $y$ by ' $t$ ' we get $\frac{dy}{dt}$.

But sometimes it is not possible to eliminate $t$, then in that case use

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

For example

If $x=a(1-\cos \theta )$ and $y=a(\theta -\sin \theta )$, then $dy/dx$ is
Solution.

$\begin{array}{rl}& \text{ Given }\mathrm{x}=\mathrm{a}(1-\cos \theta )\text{ and }\mathrm{y}=\mathrm{a}(\theta +\sin \theta )\\ & \Rightarrow \frac{dx}{d\theta }=a(\sin \theta )\text{ and }\frac{dy}{d\theta }=a(1+\cos \theta )\\ & \Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{1+\cos \theta }{\sin \theta }\end{array}$
Using the circle example, we differentiate $(-\cot (t))$ concerning

$\begin{array}{rl}& (t):[\frac{d}{dt}(-\cot (t))={\csc }^2(t)]\text{ Since }(\frac{dx}{dt}=-\sin (t)),\text{ wefind }\left(\frac{d^{2}y}{d{x}^2}\right):[\frac{d^{2}y}{d{x}^2}\\ & =\frac{{\csc }^2(t)}{-\sin (t)}=-\csc (t)]\end{array}$

List of Topics According to NCERT/JEE MAIN

Importance of Class 12 Continuity and Differentiability

Continuity and differentiability have a significant weighting in the IIT JEE test, which is a national level exam for 12th grade students that aids in admission to the country's top engineering universities. It is one of the most difficult exams in the country, and it has a significant impact on students' futures. Several students begin studying as early as Class 11 in order to pass this test. When it comes to math, the significance of these chapters cannot be overstated due to their great weightage. You may begin and continue your studies with the standard books and these revision notes, which will ensure that you do not miss any crucial ideas and can be used to revise before any test or actual examination.

How to Study Class 12 Continuity and Differentiability?

Start preparing by understanding and practicing the concept of limits. Try to be clear on the definition of continuity and differentiability and continuity and differentiability formulas. Practice many problems from each topic for better understanding. Practice continuity and differentiability class 12 previous year questions.

If you are preparing for competitive exams then solve as many problems as you can. Do not jump on the solution right away. Remember if your basics are clear you should be able to solve any question on this topic.

NCERT Notes Subject wise link:

• NCERT Notes for class 12 Physics.
• NCERT Notes for class 12 Chemistry.
• NCERT Notes for class 12 Mathematics.
• NCERT Notes for class 12 Biology

Important Books for Class 12 Continuity and Differentiability

Start from NCERT Books, the illustration is simple and lucid. You should be able to understand most of the things. Solve all problems (including miscellaneous problem) of NCERT. If you do this, your basic level of preparation will be completed.

Then you can refer to the book Amit M Aggarwal's differential and integral calculus or Cengage Algebra Textbook by G. Tewani but make sure you follow any one of these not all. Continuity and Differentiability are explained very well in these books and there are an ample amount of questions with crystal clear concepts. Choice of reference book depends on person to person, find the book that best suits you the best, depending on how well you are clear with the concepts and the difficulty of the questions you require.

NCERT Solutions Subject wise link:

• NCERT solutions for class 12 Physics.
• NCERT solutions for class 12 Chemistry.
• NCERT solutions for class 12 Mathematics.
• NCERT solutions for class 12 Biology.

NCERT Exemplar Solutions Subject wise link:

• NCERT exemplar solutions for class 12 Physics.
• NCERT exemplar solutions for class 12 Chemistry.
• NCERT exemplar solutions for class 12 Mathematics.
• NCERT exemplar solutions for class 12 Biology.