Solved Examples on Differentiation

Example 1:

i. lim _(x→0) (tan x-sin x) / x³

ii. lim (x→0) (a²-b²)/x

iii. lim _(h→0) log (x+h) – log x/(x³h)

Solution:

The given question can be solved by two methods. We discuss both the methods one by one:

Method 1:

i. lim _(x→0) (tan x – sin x)/x³ [0/0 form]

Differentiate denominator and numerator separately (i,e, apply L' Hospital rule) 

= lim _(x→0) (sec²x – cos x)/3x² [0/0 form]

Again applying L' Hospital Rule

= lim _(x→0) (2 sec x sec x tan x + sin x )/(6x) [0/0 form]

applying L' Hospital Rule again

= lim _(x→0) (2 sec⁴x + 4 tan²x sec²x + cos x)/(6)

= (2+1)/6

= 1/2

Method 2:

Alternatively, we can also solve the above question in the following way

(tan x – sin x)/x³ = (sin x (1 – cos x))/cos x . x³

sin x / x. 2 (sin² (x/2)/(x/2)²).1/4

= 1/2

Note:

Many of the problems can be solved with or without using L'Hospital rule.However you should always try to solve a problem without using L' Hospital's Rule.

ii. lim _(x→0) (a^x - b^x)/x (0/0 form)

Using L' Hospital rule

= lim _(x→0) (a^x log_ea – b^x log_eb)/1

= log_ea- log_eb

= log_e(a/b)

iii. lim _(h→0) log (x + h) – log x/h (0/0 form)

Applying L' Hospital Rule

lim _(h→0) (1/(x + h) – 0)/1 = 1/x [differentiating numerator and denominator with respect to h]

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Example 2:

(i) If g(x) = -√(25 - x²) find lim _(x→1) (g(x) – g(1))/(x-1)

(ii) Find value of a, b, c so that lim _(x→0) (ae^x - b cos x + ce^(-x))/(x sin x) = 2

Solution:

(i) lim _(x→1) (g(x) – g(1))/(x-1)

= lim _(x→1) g'(x) - g'(1)

But, g' (x) = -1/(2√(25 – x²)) (-2x)

⇒ g' (1) = 1/√24

(ii) lim _(x→0) (ae^x - b cos x + ce^(-x))/(x sin x) = 2

Let f(x) = ae^x - b cos x + ce^{- x}

And g(x) = x sin x

lim _(x→0) g(x) = 0

lim _(x→0) (ae^x -b cos x + ce^(-x))/(x sin x) = 2

So, f(x) should be zero for finite limit of f(x)/g(x)

⇒ a + c = b ....... (i)

⇒ L = lim _(x→0) (ae^x - b cos x + ce^(-x))/(x sin x) [0/0 form]

Again using L' Hospital's Rule

= lim _(x→0) (ae^x + b sin x - ce^(-x))/(x cos x + sin x)

Denominator = lim _(x→0) x cos x + sin x = 0

So for finite, lim _(x→0) ae^x + b sin x – ce^(-x) = 0

⇒ a - c = 0

a = c ........ (ii)

L = lim _(x→0) (ae^x + (2a) sin x – ae^(-x))/(x cos x + sin x ) [0/0 form] (using (i) and (ii))

   = lim _(x→0) (a(e^x + e^(-x)) + 2a cos x)/(-x sin x + 2 cos x) (applying L' Hospital rule)

   = lim _(x→0) (a(e^x + e^(-x)) + 2a cos x)/(-x sin x + 2 cos x)

   = (2a + 2a)/2

   = 2a

Given the value of limit is 2

Hence 2a = 2

⇒ a = 1

⇒ c = a = 1 (from equation (ii))

⇒ b = a + c = 1 + 1 = 2 (from equation (i))

⇒ a = 1, b = 2, c = 1.

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Example 3:

Find differential coefficient of 'y' with respect to 'x'.

1. y = 1/log cos x

2. 8^x/x⁸

3. y = tan^-1(√(1 + x² - x))

Solution:

(i) y = 1/log cos x

Let log cos x = μ

y = 1/μ (differentiating w.r.t. μ)

dy/dμ = – 1/μ²

μ = log cos x

Let cos x = t

dμ/dt = 1/t

dμ/dx = dμ/dt. dt/dx

         = 1/t (-sin x) = sin x/cos x

         = -tan x

dy/dx = dy/dμ.dμ/ dx

          = -1/μ² (-tan x)

         = tan x/(log cos x)²

Method 2:

y = (log cos x)^-1

dy/dx = (-1) (log cos x)^(-1-1). d/dx (log cos x)

         = (-1)/(log cos x)² × 1/cos x × d/dx(cos x)

         = (-1)/(log cos x)².1/cos x (-sin x)

         = tan x / (log cos x)²

Note:

Chain rule has been applied in the above question.

(2) y = 8^x/x⁸

Taking log_e on both side

log y = log 8^x - log x⁸

log y = x log 8 - 8 log x

Differentiating w.r.t. x

1/y dy/dx = log8 – 8/x

dy/dx = 8^x/x⁸ [log 8 – 8/x ]

dy/dx = 8x/(x⁹) [x log 8 - 8]

(3) y = tan^-1(√(1 + x²- x))

Subtracting x = tan θ

y = tan^-1(√(1 + x² - θ) – tan θ)

  = tan^-1(sec θ – tan θ)

  = tan^-1((1 - cos (π/2-θ))/sin (π/2-θ))

  = tan^-1(tan 1/2 (π/2-θ))

  = π/4 – 0/2 y

  = π/2-1/2 tan^-1x; by putting the value of θ

Differentiating w.r.t. x, we get

dy/dx = -1/(2(1 + x²))

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Example 4:

lim _(x→0) sin (π cos²x)/ x² equals (IIT JEE 2001)

(a) – π                                                                        (b) π

(c)  π/2                                                                       (d) 1

Solution:

lim _(x→0) sin (π cos²x)/ x² 

 = lim _(x→0) sin (π – π sin²x)/ x² 

 =  lim _(x→0) sin (π sin²x)/ (π sin²x) . (π sin²x)/(πx²) . π

 = 1.1.π

 = π

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Example 5:

If y is a function of x and log(x + y) = 2xy, then the value of y’(0) = ? (IIT JEE 2004)

(a) 1                                                                        (b) -1

(c) 2                                                                        (d) 0

Solution:

Gievn that log (x + y) = 2xy

Hence, at x = 0 we have log (y) = 0

This gives y = 1.

Now, to find dy/dx at (0, 1),

On differentiating the given equation with respect to x we have,

1/(x+y) . (1+ dy/dx) = 2x dy/dx + 2y.1

Hence, dy/dx = [2y(x + y) – 1]/[1-2(x + y)x]

So, (dy/dx)|_(0,1) = 1.

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Example 6:

If 2^x + 2^y = 2^x+y then dy/dx has the value equal to

(a) -2^y/2^x                                                                       (b) 1/(1 – 2^x)

(c) 1 – 2^y                                                                        (d) 2^x/2^y (1 – 2^y)/(2^x -1)

Solution:

The given function is 2^x + 2^y = 2^x+y

Differentiating both sides we get

2^x ln 2 + 2^y ln 2 dy/dx = 2^x+y ln 2 (1 + dy/dx)

Hence, dy/dx ( 2^x+y - 2^y) = 2^x - 2^x+y

This gives dy/dx = -2^y/2^x 

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Example 7:

Find the derivative of

1. log₇(log₇x).

2. y = cos^-1[(2 cos x + 3 sin x)/√13]

Solution:

(i) y = log₇(log₇x)

log₇x = (log_ex)/(log_e7)

Caution: Since all results in differentiation are to the base 'e'

log₇ [(log_ex)/(log_e7)] = log_e[(log_ex)/(log_e7)] / (log_e7)

(log_e7) is treated as constant with respect to x.

dy/dx  = 1/ (log_e7) . [1/(log_ex) . 1/x]

dy/dx  = 1/ (log_e7). x .log_ex

(ii) y = cos^-1[(2 cos x + 3 sin x)/√13]

= cos^-1 [2/√13 cos x + 3/√13 sin x] where 2/√13 = cos θ and 3/√13 = sin θ

= cos^-1 [cos θ cos x + sin θ sin x]

= cos^-1[cos (x – θ)]

Result: We know that cos^-1 (cos x) = x]

Hence, we have y = x - θ ; θ is a constant w.r.t 'x'

dy/dx = 1 - 0 = 1.

Note:

This example shows the significance of simplification in solving problems. If our approache had been to differentiate directly we would have never been able to get the result so quickly. Remember your approach should always be toward getting quick and correct results.

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Example 8:

If y = x log (x /(a + bx)), then show that x³(d²y)/(dx²) = [x dy/dx - y]².

Solution:

Method 1:

y = x log x – x log (a + bx)

dy/dx = x/x + 1.log x – x(b)/(a + bx) -1.log (a + bx)

          = 1 - bx/(a + bx) + log x - log (a + bx)

dy/dx = a/(a + bx) + y/x (using the given condition)

x dy/dx = ax/(a + bx) + y ........ (i)

x.(d²y)/(dx²) + 1.dy/dx = a[((a + bx) - x(b))/(a + bx)²] + dy/dx   

                                    = a²/(a + bx)² 

                                    = [(x dy/dx – y)/x]² from equation (i)

x³(d²y)/(dx²) = [x dy/dx - y]²  (Hence Proved)

Note: It is important to note the simplification of the form of dy/dx without which proof would have not been that easy.

Method 2:

y/x = log x – log (a + bx)

Differentiating with respect to x, we get

y/x = log x – log(a + bx)

Differentiating w.r.t. x

(xy' – y)/x² = 1/x – b/(a + bx)

                  = a / (x(a + bx))

⇒ xy' - y = ax/(a + bx)

Differentiating w.r.t. x again, we get

xy" + y' - y' = (a/(a + bx))²

⇒ xy" = ((xy' – y)/ x)²(using (i))

⇒ x³y" = (xy'-y)². Hence proved.

Related Resources:

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