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How to find Derivatives?

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  • Last Updated : 08 Mar, 2022
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In daily lives, people are often confronted with issues such as, “How does the patient respond to the provided doses?” or “how does the profit alter in relation to the product’s production cost?” or “What is the rate of change in pressure of a gas in relation to its volume?” or “What is the rate of change of velocity in relation to time?” and many more. If one observes carefully one can see that these questions are related to the rate of change of one varying quantity with respect to the corresponding change in another. Such a rate measure is called a derivative.

The derivative may also be used to calculate the equation of tangent and normal to the curve at a given location, as well as the maximum and minimum values of a function and its estimated values. Example,

  • A biologist uses derivatives to calculate the rate of bacterial growth in culture.
  • It is used by an electrical engineer to describe the change in current in an electric circuit.

Derivative at a point

Suppose a function f(x) is defined in a neighborhood of a.

If a and a + h belong to the domain set of a function f and 

\lim\;_{h \to 0}\;\frac{f(a + h) - f(a)}{h}

exists, then this limit is called the derivative of f(x) at x = a and is denoted by,

f‘ (a) or [df / dx]x = a

Thus, 

f'(a) = \lim\;_{h \to a}\; \frac{f(a + h) - f(a)}{h}

In general, the derivative of a function at any point x is given by,

f'(x) = \lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h} = \frac{df}{dx}

This is known as first principle of derivative.

Note: If the derivative of f(x) exists then the function f is said to be derivable or differentiable. Differentiation is the process of finding derivatives.

Geometrical Interpretation of Derivative

Consider the below graph,

Consider the function f(x) defined on an open interval (a, b). Let P be a point on the curve y = f(x). Let Q[(c – h), f(c – h)] and R be the point on either side of point P. Now,

Slope of chord PQ is, f(c – h) – f(c) / (-h).

The slope of chord PR can be written as, f(c + h) – f(c) / h

Now, we know that tangent to a curve at a point P is the limiting position of secant PQ when Q tends to P.

Similarly, it is also the limiting position of secant PR when R tends to P.

∴ as h⇢ 0 points Q and R both tend to P from left and right hand sides.

∴ The slope of tangent at point P is,

P = \lim \;_{Q \to P} \;(Slope\ of\ secant\ PQ)

\lim \;_{h \to 0} \;(Slope\ of\ secant\ PQ)

\lim \;_{Q \to P} \; \frac{f(c + h) - f(c)}{h}

If these limits exist and are equal, there is a unique tangent at point P.

The slope of tangent is denoted by dy / dx i.e., f'(x)

Thus,

f'(x) = \lim\;_{h \to 0} \; \frac{f(c + h) - f(c)}{h}

= \lim\;_{h \to 0} \; \frac{f(c - h) - f(c)}{-h}

Derivative of a standard functions

  • Constant Function

Let f(x) = k where k is any constant

∴ f(x + h) = k

Consider,

\lim\;_{h \to 0}\; \frac{f(x + h) - f(x)}{h}

=\lim\;_{h \to 0}\; \frac{k - k}{h}

= 0

∴ d(k) / dx = 0

  • Power function

Let f(x) = xn

f(x + h) = (x + h)n , n∈ R

Consider,

\lim\;_{h \to 0}\; \frac{f(x + h) - f(x)}{h}

=\lim\;_{h \to 0}\; \frac{(x + h)^n - x^n}{h}

= \lim\;_{h \to 0}\; \frac{[x^n \;+ \; nx^{n - 1}h \; + \; \frac{n(n-1)}{2!} x^{x-2}h^2 \; + .... + h^n] - x^n}{h}

= \lim\;_{h \to 0}\;\frac{h}{h}[nx^{n-1}\;+\;\frac{(n-1)}{2!}x^{n-2}h\;+......+\;h^{n-1}]

= \lim\;_{h \to 0}\; [nx^{n-1}\;+\;\frac{n(n-1)}{2!}x^{n-2}h\;+.....+h^{n-1}]

=[ nxn-1 + 0 + …….+ 0]

= n xn-1

∴ d(xn) / dx = n xn-1

Note: d(xk) / dx = kxk-1, for any real number k

  • Trigonometric Functions

Let f(x) = sin x

f(x + h) = sin(x + h)

Here,

\lim\;_{h \to 0}\; \frac{f(x + h) - f(x)}{h}

= \lim\;_{h \to 0}\; \frac {sin(x + h) - sin \;x}{h}

=\lim\;_{h \to 0}\; \frac{2cos(\frac{x + h + x}{2})sin\frac{x + h -x}{2}}{h}

=\frac{2}{2}\lim\;_{h \to 0} \;cos(\frac{2x + h}{2}) \lim\;_{h \to 0}\; \frac{sin \;h/2}{h/2})

= cos (x + 0).1

= cos x

∴ d(sin x) / dx = cos x

Similarly,

  • d(cos x) / dx = – sin x
  • d(tan x) / dx = sec2 x
  • d(cot x) / dx = – cosec2 x
  • d(sec x) / dx = sec x . tan x
  • d(cosec x) / dx = – cosec x . cot x
  • Exponential and Logarithmic function
  1. d(ax) / dx = ax . log a {a > 0 and a ≠ 1}
  2. d(ex) / dx = ex
  3. d(log x) / dx = 1 / x

Rules of Differentiation

  • Derivative of sum (Theorem 1)

Statement: If u and v are differentiable functions of x and if y = u + v then dy / dx = du / dx + dv / dx.

Proof: 

Let δu, δv, and δy be the small increments in u, v, y respectively, corresponding to the increment δx in x

As δx ⇢ 0, δu⇢ 0, δv⇢ 0, δy ⇢ 0,

Since u and v are differentiable functions of x,

\frac{du}{dx} = \lim\;_{\delta x \to0}\;\frac{\delta u}{\delta x} \; and \; \frac{dv}{dx}\lim\;_{\delta x \to0}\;\frac{\delta v}{\delta x}

y = u + v ⇢ eq (1)

y + δy = (u + δu) + (v + δv) ⇢ eq (2)

Subtracting (1) from (2) we get,

δy = δu + δv

Dividing both sides by δx, 

δy / δx = δu / δx + δv / δx

Taking the limit as δx ⇢ 0, on both sides, 

\lim\;_{\delta x \to 0}\;\frac{\delta y}{\delta x} = \;\lim\;_{\delta x \to 0}\;[\frac{\delta u}{\delta x} \;+\;\frac{\delta v}{\delta x} ]

\lim\;_{\delta x \to 0}\;\frac{\delta y}{\delta x} = \;\lim\;_{\delta x \to 0}\;\frac{\delta u}{\delta x} \;+\lim\;_{\delta x \to 0}\;\frac{\delta v}{\delta x}

du / dx + dv / dx    

\therefore \lim\;_{\delta x\to0}\; \frac{\delta y}{\delta x }

∴ dy / dx = du / dx + dv / dx

Similarly,

  • Derivative of difference (Theorem 2)

Statement: If u and v are differentiable functions of x and if y = u – v then dy / dx = du / dx – dv / dx.

  • Derivative of Product (Theorem 3)

Statement: If u and v are differentiable functions of x and if y = u  v then 

\frac{dy}{dx} = u\;\frac{dv}{dx}\; + \; v \;\frac{du}{dx}

Proof:

Let δu, δv, and δy be the small increments in u, v, y respectively, corresponding to the increment δx in x

As δx ⇢ 0, δu⇢ 0, δv⇢ 0, δy ⇢ 0,

Since u and v are differentiable functions of x,

\frac{du}{dx} = \lim\;_{\delta x \to0}\;\frac{\delta u}{\delta x} \; and \; \frac{dv}{dx}\lim\;_{\delta x \to0}\;\frac{\delta v}{\delta x}

y = uv ⇢ eq. (1)

y + δy = (u + δu) (v + δv)        

= uv + uδv + vδu + δuδv ⇢ eq(2)

Subtracting (1) from (2), we get

δy = uδv + vδu + δuδv

Dividing both the sides by δx, 

δy / δx = u(δv / δx) + v(δu / δx) + (δu / δx)δv

Taking limit as δx ⇢ 0, we get

\lim\;_{\delta x \to 0}\; \frac{\delta y}{\delta x} = u\lim\;_{\delta x \to 0}\; \frac{\delta v}{\delta x}\; + \; v\lim\;_{\delta x \to 0}\; \frac{\delta u}{\delta x}\; + \; \lim\;_{\delta x \to 0}\;\frac{\delta u}{\delta x}.\lim\;_{\delta x\to 0}\;\delta v

= u (dv / dx) + v(du / dx) + (du/dx) × 0

As R.H.S. exists and is equal to (dy / dx),

\frac{dy}{dx} = u\;\frac{dv}{dx}\; + \; v \;\frac{du}{dx}

Thus, the derivative of product of two functions = first function × derivative of second function + second function × derivative of first function

Similarly,

  • Derivative of quotient (Theorem 4)

Statement:  If u and v are differentiable functions of x and if y = u / v then 

\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

Sample problems

Question 1: Find the derivative of 3x + 4 using the first principle of derivative.

Solution:

Let f(x) = 3x + 4

f(x + h) = 3(x + h) + 4

consider,

\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}

\lim\;_{h \to 0} \; \frac{3(x + h)\; + \; 4 \; - \;(3x + 4)}{h}

\lim\;_{h \to 0}\; \frac{3h}{h}

= 3

∴ f’ (x) = 3

Question 2: Find the derivative of

  • 1 / x2 and
  • x cos x.

Using the first principle of derivative.

Solution:

  • 1 / x2

Let f(x) = 1 / x2

f(x + h) = 1 / (x + h)2

Consider,

\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}

\lim\;_{h \to 0} \; \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}

\lim\;_{h \to 0}\; \frac{x^2 - (x + h^2)}{h(x + h)^2 . x^2}

\lim\;_{h \to 0}\; \frac{x^2 - (x^2 + 2xh + h^2)}{h(x + h)^2 . x^2}

\lim\;_{h \to 0}\; \frac{-h(2x + h)}{h(x + h)^2 . x^2}

-\lim\;_{h \to 0}\; (2x + h)\;\;\lim\;_{h \to 0}\; \frac{1}{(x + h)^2 . x^2}

= -2x(1 / x2x2)

= -2 / x3

∴ f'(x) = -2 / x3

  • x cos x

Let f(x) = x cos x

f(x + h) = (x + h) cos (x + h)

consider,

\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}

\lim\;_{h \to 0} \; \frac{(x + h) cos (x + h) - x cos x}{h}

\lim\;_{h \to 0} \; \frac{x[cos (x + h) - cos x]+ h\;cos(x + h)}{h}

\lim\;_{h \to 0} \; \frac{x(-2)sin(x + \frac{h}{2})\;sin\frac{h}{2}}{2x\frac{h}{2}}\; + \; \lim\;_{h \to 0}\; \frac{h\;cos(x + h)}{h}

-x\;\lim\;_{h \to 0}\; sin (x +\frac{h}{2}).\; \lim\;_{h \to 0}\; \frac{sin\frac{h}{2}}{\frac{h}{2}}\;+\; cos(x + 0)

= -x sin x. 1 + cos x

= -x sin x + cos x

∴ f'(x) = -x sin x + cos x

Question 3: Differentiate the following

  • sin x / 1 + sin x
  • ex / 1 + sin x

Solution:

  • sin x / 1 + sin x

we have

y = sin x / 1 + sin x

\frac{dy}{dx} = \frac{(1 + sin\;x) \frac{d}{dx}(sin\;x) - sin\;x \frac{d}{dx}(1 + sin\;x)}{(1+ sinx)^2}

=\frac{(1 + sin\;x).cos\;x  - sin\;x. (0 + cos\;x)}{(1+ sinx)^2}

=\frac{cos\;x + sin\;x.cos\;x - sin\;x.cos\;x}{(1 + sin\;x)^2}

= cos x / (1 + sin x)2

  • ex / 1 + sin x

y = ex / 1 + sin x

\frac{dy}{dx}\; = \frac{(1 + sin\;x)\frac{d}{dx}e^x\;-\;e^x\;\frac{d}{dx}(1 + sin\;x)}{(1 + sin\;x)^2}

=\frac{(1 + sin\;x).e^x\;-\;e^x.(0 + cos\;x)}{(1 + sin\;x)^2}

= \frac{e^x[1 + sin\;x - cos\;x]}{(1 + sin\;x)^2}

Question 4: y = log 5(log7 x) find dy / dx

Solution:

we have,

y  = log5 (log7x)

Using logarithmic property, we can write,

y = log( log7x ) / log 5

\frac{dy}{dx} = \frac{1}{log\;5}.\frac{d}{dx}[log(log_7x)]

= \frac{1}{log\;5}\;.\;\frac{1}{\frac{log\;x}{log\;7}}\;.\;\frac{d}{dx}(\frac{log\;x}{log\;7})

= \frac{1}{log\;5}\;.\;\frac{1}{\frac{log\;x}{log\;7}}\;.\;\frac{1}{log\;7}\frac{d}{dx}(log\;x)

= \frac{1}{log\;5\;.\;.log\;x}\;.\;\frac{1}{x}

= 1 / x log 5. log x

Question 5: If y = log[\frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25}-x}]   

Solution:

y = log[\frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25}-x}]

y = log\;(x + \sqrt{x^2 + 25})\; -\; log\;(\sqrt{x^2 + 25}-x)

\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2+25}}\;.\;\frac{d}{dx}(x + \sqrt{x^2+25})\;-\;\frac{1}{\sqrt{x^2+25}-x}\;.\;\frac{d}{dx}(\sqrt{x^2 + 25}-x)

= \frac{1}{x+\sqrt{x^2+25}}\;(1 \;+\;\frac{1}{2\sqrt{x^2+25}}\times2x)\;-\;\frac{1}{\sqrt{x^2+25}-x}\;(\frac{1\times2x}{2\sqrt{x^2+25}}\;-\;1)

= \frac{1}{x + \sqrt{x^2+25}}\;[\frac{\sqrt{x^2+25}\;+\;x}{\sqrt{x^2+25}}]\;+\;\frac{1}{\sqrt{x^2+25}-x}\;[\frac{\sqrt{x^2+25}-x}{\sqrt{x^2+25}}]

= \frac{1}{\sqrt{x^2+25}}\;+\;\frac{1}{\sqrt{x^2+25}}

= \frac{2}{\sqrt{x^2+25}}

Question 6: If y = log(3x2 + 2x +1), find dy / dx

Solution:

y = log (3x2 + 2x +1)

= \frac{dy}{dx} = \frac{1}{3x^2 + 2x + 1}\;\;\frac{d}{dx}(3x^2 + 2x + 1)

= \frac{1}{3x^2+2x+1}\;\;(6x +2)

= \frac{2(3x +1)}{3x^2 + 2x +1}

Question 7: Using the first principle of derivative find √sin x

Solution:

Let f(x) = √sin x

f(x + h) = √sin (x + h)

consider,

\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}

= \lim\;_{h \to 0}\;\frac{\sqrt{sin(x + h)} \;-\;\sqrt{sin\;x} }{h}

= \lim\;_{h \to 0}\;\frac{\sqrt{sin(x + h)} \;-\;\sqrt{sin\;x} }{h} \; \times \;= \frac{\sqrt{sin(x + h)} \;+\;\sqrt{sin\;x} }{\sqrt{sin(x + h)} \;+\;\sqrt{sin\;x}}

= \;\lim\;_{h \to 0}\; \frac{sin(x + h) - sin\;x}{h[\sqrt{sin(x + h)} \;+\; \sqrt{sin\;x} ]}

= \lim\;_{h \to 0}\; \frac{2 cos\;(\frac{x + h+ x}{2})\;.\;sin\;(\frac{x + h - x}{2})}{h[\sqrt{sin(x + h)}\; +\;\sqrt{sin\;x} ]}

= \lim\;_{h \to 0}\; \frac{2\;cos\;(x +\frac{h}{2})\;.\;sin(\frac{h}{2})}{h[\sqrt{sin(x + h)\;+\;\sqrt{sin\;x}}]}

= \lim\;_{h \to 0}\; \frac{\;cos\;(x +\frac{h}{2})}{\sqrt{sin(x + h)\;+\;\sqrt{sin\;x}}}\;\times\;\frac{sin(\frac{h}{2})}{(h/2)}

=\frac{\lim\;_{h \to 0}\;cos\;(x +\frac{h}{2})}{\lim\;_{h\to 0}\;\sqrt{sin(x + h)\;\;+\sqrt{sin\;x}}}\;\times\;\lim\;_{h \to 0}\;\frac{sin(\frac{h}{2})}{(h / 2)}

=\frac{cos\;x}{\sqrt{sin\;x}\;+\;\sqrt{sin\;x}}\;\times1

.....[h \to 0, \frac{h}{2} \to 0 \;and\;\lim\;_{\theta \to 0}\;\frac{sin\;\theta}{\theta} =1]

= \frac{cos\;x}{2\sqrt{sin\;x}}


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