Calculus (Part-2)-geometric explanation of differentiation

Online maths tutors like the following  concept very much

Geometric explanation of differentiation-  If we find the derivative of function f(x) at x = x0 then it is equal to the slope of the tangent  to the graph of given function f(x) at the given point [(x0, f(x0))].

But what is a tangent line?

It is not merely a simple line that joins the graph of given function at one point.

It is actually the limit of the secant lines joining points P = [(x0, f(x0)] and Q on the graph of f(x) as Q moves very much close to P.

The tangent line contacts the graph of given function at given point [(x0, f(x0)] the slope of the tangent line matches the direction of the graph at that point. The tangent line is the straight line that best approximates the graph at that point.

<img src="derivative and slope.png" alt="derivative and slope">


As we are given the graph of given function, we can draw the tangent to this graph easily. Still, we’ll like to make calculations involving the tangent line and so will require a calculative method to explore the tangent line.


We can easily calculate the equation of the tangent line by using the slope-point form of the line. We slope of a line is m and its passing through a point (x0,y0) then its equation will be

                                                              y − y0 = m(x − x0)


So now we have the formula for the equation of the tangent line. It’s clear that to get an actual equation for the tangent line, we should know the exact coordinates of point P. If we have the value of x0 with us we calculate y0 as

                                                                      y = f(x0)

The second thing we must know is the slope of line

                                                                      m = f’(x0)

Which we call the derivative of given function f(x).



The derivative f’(x0) of given function f at x0 is equal to the slope of the tangent line to

                                        y = f(x) at the point P = (x0, f(x0).

Differentiation Using Formulas- We can use derivatives of different types of functions to solve our problems :


{\rm{(i) D (}}{{\rm{x}}^{\rm{n}}}{\rm{) = n}}{\rm{.}}{{\rm{x}}^{{\rm{n - 1}}}}{\rm{ here n, x}} \in R        \;\left( {ii} \right){\rm{ }}\;D{\rm{ }}\left( {{e^x}} \right){\rm{ }} = {\rm{ }}{e^x}

{\rm{(iii) D(}}{{\rm{a}}^{\rm{x}}}{\rm{) = }}{{\rm{a}}^{\rm{x}}}{\rm{lo}}{{\rm{g}}_e}{\rm{a a > 0}}                {\rm{(iv) D (ln x) = }}\frac{1}{x}

{\rm{(v) D(lo}}{{\rm{g}}_{\rm{a}}}{\rm{x) = lo}}{{\rm{g}}_{\rm{a}}}{\rm{e }}                      {({\rm{vi}}){\rm{D}}({\rm{sinx}}){\rm{ = cosx}}}

{({\rm{vii}}){\rm{D}}({\rm{cosx}}){\rm{ = - sinx}}}                     {({\rm{viii}}){\rm{D(tanx) = se}}{{\rm{c}}^2}x}

(ix)  D (secx) = secx . tanx              (x)  D (cosecx) = – cosecx . cotx   

{\rm{(xi) D (cotx) = - cose}}{{\rm{c}}^2}x              (xii) D (constant) = 0 where D = {\frac{d}{{dx}}}

These formulas are result of differetiation by first principle

Inverse Functions And Their Derivatives :


{\rm{(i) }}\frac{d}{{dx}}({{{\rm Sin}\nolimits} ^{ - 1}}x) = \frac{1}{{\sqrt {{1^2} - {x^2}} }}                 {\rm{(ii) }}\frac{d}{{dx}}({\cos ^{ - 1}}x) = \frac{{ - 1}}{{\sqrt {{1^2} - {x^2}} }}

{\rm{(iii) }}\frac{d}{{dx}}({{{\rm Sec}\nolimits} ^{ - 1}}x) = \frac{1}{{x\sqrt {{x^2} - 1} }}              {\rm{(iv) }}\frac{d}{{dx}}({{{\rm cosec}\nolimits} ^{ - 1}}x) = \frac{{ - 1}}{{x\sqrt {{x^2} - 1} }}

{\rm{(v) }}\frac{d}{{dx}}(Co{t^{ - 1}}x) = \frac{{ - 1}}{{1 + {x^2}}}                  {\rm{(vi) }}\frac{d}{{dx}}({\tan ^{ - 1}}x) = \frac{1}{{1 + {x^2}}}

Theorems On Derivatives:  If u and v are derivable function of x, then,


(i)   \begin{array}{l} {\rm{ }}\frac{d}{{dx}}(u \pm v) = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}\\ \\ \end{array}                    (ii)   \frac{d}{{dx}}k[f(x)] = k\frac{d}{{dx}}f(x)    where K is any constant


(iii)  {\rm{ }}\frac{d}{{dx}}(u.v) = v\frac{{du}}{{dx}} + u\frac{{dv}}{{dx}}    known as  “ Product  Rule ”      (iv)   {\rm{ }}\frac{d}{{dx}}(\frac{u}{v}) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}

known as  “Quotient  Rule ”


(v) If  y = f(u)  &  u = g(x)  then  \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}}    “Chain  Rule ”


Logarithmic  Differentiation:  To find the derivative of  : (i) a function which is the product or quotient of a number of functions OR
(ii) a function of the form  {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} where f & g  are both differentiable, it will be found convenient to take the logarithm of the function first & then differentiate. This is called Logarithmic  Differentiation.


Implicit  Differentiation:  (i) In order to find dy/dx,  in the case of implicit functions, we differentiate each term w.r.t.  x  regarding y as a function of x & then collect terms in dy/dx together on one side to finally find dy/dx.

(ii) In answers of dy/dx in the case of implicit functions, both x & y are present.

Parametric  Differentiation: If  y = f(q)  &  x = g(q)  where q  is  a parameter, then

                                                                         \frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }}

Derivative Of A Function w.r.t. Another Function-: Let  y = f(x) & z = g(x)

\frac{{dy}}{{dx}} = \frac{{f'(x)}}{{g'(x)}}

Derivatives Of Order Two & Three: Let a function y = f(x) be defined on  an open interval (a, b).  It’s derivative, if it exists on(a, b)  is  a certain function f'(x) [or (dy/dx) or y’ ] &  is  called  the  first derivative of y  w.r.t. x. If it happens that the first derivative has a derivative on (a, b) then this derivative is called the second derivative of y  w. r. t.  x  &  is denoted by f”(x) or (d2y/dx2) or  y”. Similarly, the 3rd order derivative of y w. r. t.  X, if it exists, is defined by    It is also denoted by f”'(x) or y”’.

All online maths tutors suggest solving fair amount of questions based on these concepts


Example-Find the tangent line to the following function at z=3 for the given function


We can find the derivative of the given function using basic differentiation as discussed in the previous post            

We are already given that z=3 so

Equation of tangent line is

                                                    y − y0 = m(x − x0)      here y0=R(3)=√7

Putting these values we get equation of tangent line


In my online maths tutors series, I will discuss Application of Derivatives


Example-  Differentiate the following function    


Ans: We can apply quotient rule in this questions


You can try these questions in the meantime

 28 – Differentiation (09-10-15) with answers.pdf



<img src="demo.png" alt="demo">



How to find sum and product of zeros of equations

In my previous post on IB Mathematics, I have discussed how to solve a quadratic polynomial using Quadratic formula. Here I will tell you about different relationships based on sum and product of quadratic polynomial, cubic polynomial and bi-quadratic polynomials.

<img src="ib mathematics.png" alt="ib mathematics">


ax2 + bx + c = 0

                                         Sum of the roots = −b/a

                                          Product of the roots = c/a

If we know the sum and product of the roots/zeros of a quadratic polynomial, then we can find that polynomial using this formula

x2 − (sum of the roots)x + (product of the roots) = 0


Now let us look at a Cubic (one degree higher than Quadratic):

ax3 + bx2 + cx + d=0

 if α, β and γ are the zeros of this cubic polynomial then

If we know these relationships of polynomials then we cal also calculate the polynomial using this formula:


If we are given a bi-quadratic polynomial with degree 4 like:

                                             ax 4+bx³+cx²+dx+e=0
                                and its roots/zeros are α, β, γ, and δ then
using these formulas of sum and product of zeros of polynomials, we can find a lot of relationships in zeros of polynomials. Usually, we are asked to find these types of relationships in zeros.
Question: If α and β are the zeros of polynomial x²-px+q=0 then find the following relationships.
i) 1/α+1/β   ii) α²β+αβ²   iii) α²+β²    iv) α/β+β/α   v) α³+β³
Ans: To find these relationships we convert every value either in sum (α+β)  or in the product (αβ) of zeros. For this conversion, we use following Mathematical Tricks
1)Try to take common
2) Try to take L.C.M
3) Try to make a Perfect Square
4) Use algebraic identities wherever required
If we use above steps properly, in most of the cases we are able to convert everything either in sum or in product of zeros
If we compare the given equation with std. form ax²+bx+c=0 then
                                                                       a=1, b=-p and c=q
                             sum of zeros α+β=-b/a=-(-p/a)=p

                             product of zeros    αβ=c/a=q/1=q                                                     (i) 1/α+1/β=β+α/αβ              [By L.C.M]

(ii) α²β+αβ²  = αβ(α+β)               [By common]
 (iii) α²+β² = (α²+β²+2αβ)-2αβ          [add and subtract 2αβ, make it a perfect square ]
(iv) α/β+β/α = β²+α²/βα          [By taking L.C.M]
we have already found β²+α² that is p²-2q so

(v) (α+β)³=(α+β)³-3αβ(α+β)                [Direct algebraic identity]

You can further read about quadratics in PDF(quadratics ) given here. There are a lot of practise questions given in this PDF
<img src="demo.png" alt="demo">