Mathematics

### Topics Covered

color{red} star Derivatives - Definition
color{red} star Algebra of derivative of functions
color{red} star Derivative of f(x) = x^n is nx^(n – 1) for any positive integer n.
color{red} star Derivative of polynomial functions

### Derivatives

●\color{fuchsia} {ul"Definition 1 : "} Suppose f is a real valued function and a is a point in its domain of definition. The derivative of f at a is defined by

color(blue)(lim_(h →0) ( f(a+h) - f(a))/h) provided this limit exists.

Derivative of f (x) at a is denoted by f′(a). Observe that f′ (a) quantifies the change in f(x) at a with respect to x.

Q 3121612521

Find the derivative at x = 2 of the function f(x) = 3x.

Solution:

We have

f'(2) = lim_(h→0) (f(2+h)-f(2))/(h)= lim_(h→0) (3(2+h)-3(2))/(h)

lim_(h→0) (6+3h-6)/(h) = lim_(h→0)(3h)/(h) = lim_(h→0) 3 = 3

The derivative of the function 3x at x = 2 is 3.
Q 3150812714

Find the derivative of the function f(x) = 2x^2 + 3x – 5 at x = –1. Also prove that f ′ (0) + 3f′ ( –1) = 0.

Solution:

We first find the derivatives of f(x) at x = –1 and at x = 0. We have

f' (-1) = lim_(h→0) (f(-1+h)-f(-1))/h

= lim_(h→0) ([ 2(-1+h)^2+3(-1+h)-5]-[2(-1)^2+3(-1)-5])/h

= lim_(h→0) (2h^2-h)/h = lim_(h→0) (2h-1) = 2(0) -1 = -1

and f'(0) = lim_(h→0) (f(0+h)-f(0))/h

 = lim_(h→0) ( [ 2(0+h)^2+3(0+h)-5] - [ 2(0)^2+3(0) -5])/h

 = lim_(h→0) (2h^2+3h)/h = lim_(2h+3) = 2(0)+3 = 3

f'(0)+3f'(-1) = 0
Q 3161612525

Find the derivative of sin x at x = 0.

Solution:

Let f(x) = sin x. Then

f'(0) = lim_(h→0) (f(0+h) -f(0))/(h)

= lim_(h→0) (sin(0+h) -sin (0)) /(h) = lim_(h→0) (sin h)/(h) = 1
Q 3171612526

Find the derivative of f(x) = 3 " at " x = 0 and at x = 3.

Solution:

Since the derivative measures the change in function, intuitively it is clear that the derivative of the constant function must be zero at every point. This is indeed, supported by the following computation.

f'(0) = lim_(h→0)(f(0+h)-f(0))/(h)= lim_(h→0)(3-3)/(h) = lim_(h→0)(0)/(h) = 0

Similarly f'(3) = lim_(h→0)(f(3+h) - f(3))/(h) lim_(h→0)(3-3)/(h)=0.

We now present a geometric interpretation of derivative of a function at a point. Let y = f(x) be a function and let P = (a, f(a)) and Q = (a + h, f(a + h) be two points close to each other on the graph of this function. The Fig 13.11 is now self explanatory.

We know that f'(a) = lim_(h→0) (f(a+h)-f(a))/(h)

From the triangle PQR, it is clear that the ratio whose limit we are taking is precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process, as h tends to 0, the point Q tends to P and we have

= lim_(h→0) (f(a+h)-f(a))/(h) = = lim_(Q→P) (QR)/(PR)

= lim_(h→0) (f(a+h)-f(a))/(h)== lim_(Q→P) (QR)/(PR)

This is equivalent to the fact that the chord PQ tends to the tangent at P of the curve y = f(x). Thus the limit turns out to be equal to the slope of the tangent. Hence
f ′(a) = "tan" ψ .
For a given function f we can find the derivative at every point. If the derivative exists at every point, it defines a new function called the derivative of f .

### Definition 2 :

\color{fuchsia} {ul"Definition 2 : "} Suppose f is a real valued function, the function defined by

color(blue)(lim_(h→0) ( f(x+h) - f(x))/h)

wherever the limit exists is defined to be the derivative of f at x and is denoted by f′(x).

color(red)("This definition of derivative is also called the first principle of derivative.")

Thus color(blue)(f'(x) = lim_(h→0) (f (x+h) - f(x))/h)

Clearly the domain of definition of f′ (x) is wherever the above limit exists.

There are color(green)("different notations for derivative of a function.")

\color{green} ✍️  Sometimes f′ (x) is denoted by color(blue)(d/(dx) ( f(x))) or if y = f(x). it is denoted by color(blue)((dy)/(dx)). This is referred to as derivative of f(x) or y with respect to x.

\color{green} ✍️ It is also denoted by color(blue)(D (f (x) )). Further, derivative of f at x = a is also denoted by d/(dx) f(x) |_(a)  or (df)/(dx) |_(a) or even ((df)/(dx))_(x= a)
Q 3181612527

Find the derivative of f(x) = 10x.

Solution:

Since f′ ( x) =lim_(h→0) (f(x+h)-f(x))/(h)

lim_(h→0)(10(x+h)-10(x))/(h)

lim_(h→0) (10h)/(h) = lim_(h→0) (10)=10
Q 3111612529

Find the derivative of f(x) = x^2.

Solution:

We have, f ′(x) lim_(h→0) (f(x+h)-f(x))/(h)

lim_(h→0)((x+h)^2-(x)^2)/(h) = lim_(h→0) (h+2x)=2x
Q 3111712620

Find the derivative of the constant function f (x) = a for a fixed real number a.

Solution:

We have, f ′(x) =lim_(h→0) (f(x+h)-f(x))/(h)

lim_(h→0) (a-a)/(h) = lim_(h→0) (0)/(h) =0 "as" h ≠ 0
Q 3121712621

Find the derivative of f(x) = 1/x

Solution:

We have f ′(x) = lim_(h→0) (f(x+h)-f(x))/(h)

= lim_(h→0)(1/(x+h)-1/x)/h

= lim_(h→0)1/h[(x-(x+h))/(x(x+h))]

= lim_(h→0)1/h[(-h)/(x(x+h))] = = lim_(h→0) (-1)/(x(x+h)) = -1/(x^2)

### Algebra of derivative of functions

● Since the very definition of derivatives involve limits in a rather direct fashion, we expect the rules for derivatives to follow closely that of limits. We collect these in the following theorem.

● Let f and g be two functions such that their derivatives are defined in a common domain. Then
.
color(red)((i)) Derivative of sum of two functions is sum of the derivatives of the functions.

color(blue)[d/(dx) { f(x) + g(x) } = d/(dx) f(x) +d/(dx) g (x)].

color(red)((ii)) Derivative of difference of two functions is difference of the derivatives of the functions.

color(blue)[d/(dx) { f(x) - g(x) } = d/(dx) f(x) - d/(dx) g(x)].

color(red)((iii)) Derivative of product of two functions is given by the following color(red)"product rule."

color(blue)[d/(dx) { f(x) * g(x) } = d/(dx) f(x) * g(x) + f(x) * d/(dx) g(x)]

color(red)((iv)) Derivative of quotient of two functions is given by the following color(red)"quotient rule" (whenever the denominator is non–zero).

color(blue)[d/(dx) ((f(x))/(g(x))) = ( d/(dx) f(x) * g(x) - f(x) d/(dx) g(x))/(g(x))^2]

Proof : Let color(red)(u = f (x) and v = g (x).) Then

color(green)((uv)′ = u′v + uv′)
This is referred to a Leibnitz rule for differentiating product of functions or the product rule. Similarly, the quotient rule is

color(orange)((u/v)' = (u'v-uv')/(v^2))

Now, let us tackle derivatives of some standard functions. It is easy to see that the derivative of the function f(x) = x is the constant

function 1. This is because color(blue)(f '(x)=lim_(h →0) (f(x+h)-f(x))/h = lim_(h →0)(x+h-x)/h

color(red)(lim_(h →0)1=1)

We use this and the above theorem to compute the derivative of

color(blue)(f(x) = 10x = x + .... + x) (ten terms). By (i) of the above theorem

color(red)((df)/(dx) = (d)/(dx) (x+....+x)) (ten terms)

color(orange)(= d/(dx)x+..+ d/(dx) x) (ten terms)
color(navy)(= 1+ ... +1 (ten terms) = 10.)

● We note that this limit may be evaluated using product rule too. Write f(x) = 10x = uv, where u is the constant function taking value 10 everywhere and v(x) = x. Here, f(x) = 10x = uv we know that the derivative of u equals 0. Also derivative of v(x) = x equals 1. Thus by the product rule we have

color(blue)(f ′(x) = (10x)′ = (uv)′ = u′v + uv′ = 0.x +10.1 =10)

● On similar lines the derivative of f(x) = x^2 may be evaluated. We have f(x) = x^2 = x .x and hence

color(green)((df)/(dx) = d/(dx) (x.x) = d/(dx) (x).x+x.d/(dx)(x))

color(blue)(l=1x+x.1=2x

More generally, we have the following theorem.

color{blue} {"Theorem 6 :"} Derivative of f(x) = x^n is nx^(n – 1) for any positive integer n.

color(red)("Proof:") By definition of the derivative function, we have

color(green)(f'(x) = lim_(h →0) (f (x+h) - f(x))/h = lim_( h→0) (( x+h)^n-x^n)/h)

Binomial theorem tells that color(purple)((x+h)^n = ( text()^(n)C_0) x^n +( text()^(n)C_1) x^(n-1) h +...........+ ( text()^(n)C_n) h^n)

and hence color(purple)((x+h)^n - x^n = h( n x^(n-1) +........+ h^(n-1)) ) .

Thus color(blue)(( df (x))/(dx) = lim_(h→0) (( x+h)^n - x^n)/h)

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = lim_(h →0) ( h ( n x^(n-1) + .......+ h^(n-1)))/h

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = lim_( h → 0) ( n x^(n-1) +........+ h^(n-1)) = color(red)(nx^(n-1))

color(green)"Alternatively," we may also prove this by induction on n and the color(blue)"product rule" as follows.
We have

color(blue)(d/(dx) (x^n) = d/(dx) ( x . x^(n-1)))

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= d/(dx) (x) . (x^(n-1)) + x. d/(dx) (x^(n-1)) (by product rule)

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 1 . x^(n-1) + x . ( (n-1) x^(n-2)) (by induction hypothesis)

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = x^(n-1) + (n-1) x^(n-1) = color(red)(n x^(n-1))

Remark : The above theorem is true for all powers of x, i.e., n can be any real number (but we will not prove it here).

### Derivative of polynomials and trigonometric functions

We start with the following theorem which tells us the derivative of a polynomial function.

color(red)(ul"Theorem : 7 ") Let f(x) = a_n x^n +a_(n-1) x^(n-1) +...........+ a_1 x+a_0 be a polynomial function, where a_1 are all real numbers and a_n ≠ 0.

Then, the derivative function is given by

color(blue)((df(x))/(dx) = n a_n x^(n-1) + (n-1) a_(n-1) x^(x-2) +.............+ 2a_2x+a_1)
Q 3131712622

Compute the derivative of 6x^100 – x^55 + x.

Solution:

A direct application of the above theorem tells that the derivative of the above function is 600x^99 − 55x^54 +1.
Q 3141712623

Find the derivative of f(x) = 1 + x + x^2 + x^3 +... + x^50 at x = 1.

Solution:

direct application of the above Theorem tells that the derivative of the above function is

1 + 2x + 3x^2 + . . . + 50x^49.  At x = 1 the value of this function equals

1 + 2(1) + 3(1)^2 + .. . + 50(1)^49

= 1 + 2 + 3 + . . . + 50 = ((50)(51))/(2) = 1275.
Q 3151712624

Find the derivative of f(x) = (x+1)/(x)

Solution:

Clearly this function is defined everywhere except at x = 0.

We use the quotient rule with u = x + 1 and v = x. Hence u′ = 1 and v′ = 1.

Therefore (df(x))/(dx) = d/(dx) (x+1)/(x) = d/(dx) (u/v)

" "= (u'v-uv')/(v^2) = (1(x) - ((x+1)1)) /(x^2) = -1/(x^2)
Q 3161712625

Compute the derivative of sin x.

Solution:

Let f(x) = sin x. Then

(df(x))/(dx) = lim_(h→0) (f(x+h)-f(x))/(h)= lim_(h→0) (sin (x+h)- sin (x))/(h)

lim_(h→0) (2cos((2x+h)/(2)) sin (h/2))/(h) (using formula for sin A – sin B)

lim _(h→0) cos(x+h/2) lim_(h→0) (sin (h/2))/(h/2) = cos x. 1 = cos x
Q 3171712626

Compute the derivative of tan x.

Solution:

Let f(x) = tan x. Then
(df(x))/(dx) = lim_(h→0) f(x+h)-f(x)/(h) = lim_(h→0) (tan(x+h)-tan (x)/(h)

lim_(h→0)1/h [ (sin(x+h))/(cos (x+h) - (sinx)/(cosx)]]

lim_(h→0)[ (sin (x+h) cosx - cos(x+h) sin x)/(hcos(x+h) cosx)]

lim_(h→0)(sin (x+h-x))/cos(x+h)cosx (using formula for sin (A + B)

lim_(h→0) (sinh)/(h) lim_(h→0) (1)/(cosx+h)cos x

= 1 (1)/(cos^2x) = sec^2x
Q 3181712627

Compute the derivative of f(x) = sin^2 x.

Solution:

We use the Leibnitz product rule to evaluate this.

(df(x))/(dx) = d/(dx) (sinx sin x)

= (sin x)' sin x+sin x(sin x)'

= (cosx) sin x+ sin (cosx)

= 2sin x cosx= sin 2x.