Samacheer Class 11 Maths - Differentiability and Methods of Differentiation

(i) f(x) = 6
(ii) f(x) = – 4x + 7,
f(x + Δx) = -4(x + Δx) + 7
f(x + Δx) – f(x) = [-4(x + Δx) + 7] – [-4x + 7]
f(x + Δx) – f(x) = [-4(x + Δx) + 7] + 4x – 7
f(x + Δx) – f(x) = -4 Δx
(iii) f(x) = -x 2 + 2
f (x + Δx) = – (x + Δx) 2 + 2
f (x + Δx) – f(x) = – [x 2 + 2x Δx + (Δx) 2 ] + 2 – [- x 2 + 2]

(ii) f(x) = |x 2 – 1|at x = 1
(iii) f(x) = |x| + |x – 1| at x = 0, 1
To find the limit at x = 0
First we find the left limit of f(x) at x = 0
When x = 0 – |x| = -x and
|x – 1| = -(x – 1)
∴ When x = 0 we have
f(x) = -x – (x – 1)
f(x) = -x – x + 1 = -2x + 1
f(0) = 2 × 0 + 1 = 1
f'(0 – = – 2 ……… (1)
∴When x = 0 + we have
|x| = x and |x – 1| = – (x – 1)
∴ f(x) = x – (x – 1)
f(x) = x – x + 1
f(x) = 1
f(0) = 1
From equations (1) and (2), we get
f'(0 – ) ≠ f’(0 + )
∴ f(x) is not differentiable at x = 0.
To find the limit at x = 1
First we find the left limit of f (x) at x = 1
When x = 1, |x| = x and
|x – 1| = -(x – 1)
∴ f(x) = x – (x – 1)
f(x) = x – x + 1 = 1
f(x) = 1
f(1) = 1
When x = 1 +, |x| = x and
|x – 1| = x – 1
When x =, |x| = x and
|x – 1| = x – 1
∴ f(x) = x + x – 1 = 2x – 1
f(1) = 2 × 1 – 1 = 2 – 1 = 1
From equations (3) and (4), we get
f’(1 – ) ≠ f'(1 + )
∴ f (x) is not differentiable at x = 1
(iv) f(x) = sin |x| at x = 0
First we find the left limit of f (x) at x = 0
When x = 0 –, |x| = -x
∴ f(x) = sin (-x) = -sin x
f(0) = -sin 0 = 0
Next we find the right limit of f (x) at x = 0
When x = 0 + |x| = x
∴ f(x) = sin x
f(0) = sin 0 = 0
From equations (1) and (2), we get
f’(0 – ) ≠ f'(0 + )
∴ f (x) is not differentiable at x = 0.

f(x) = |x + 100| + x 2
First let us find the left limit of f( x) at x = – 1100
When x < – 100,
f(x) = – (x + 100) + x 2
f(- 100) = – (- 100 + 100) + (- 100) 2
f(- 100) = 100 2
= -1 – 100 – 100
f’ (-100) = -201 ——– (1)
Next let us find the right limit of f( x) at x = -100
when x > – 100
f(x) = x + 100 + x 2
f(- 100) = – 100 + 100 + (- 100) 2
f(- 100) = 100 2
f'(-100 + ) = – 199 ……… (2)
From equation (1) and (2), we get
f’(- 100 – ) ≠ f'(- 100 + )
∴ f’ (x) does not exist at x = -100
Hence, f'(- 100) does not exist

At the points, x = 0, π, 2π, 3π, ……….. the graph of the given function has a sharp edge V.
∴ At these points, the function is not differentiable.
∴ The function y = |sin x| is not differentiable at
x = nπ, for all n ∈ Z.
(ii) |cos x|

f(x) = uv
⇒ f'(x) = uv’ + vu’ = u\(\frac{d u}{d x}\) + v\(\frac{d v}{d x}\)
Now u = x ⇒ u’ = 1
v = sin x ⇒ v’ cos x
f'(x) = x (cos x) + sin x(1)
= x cos x + sin x

g(t) = t 3 cost (i.e.) u = t 3 and v = cos t
let u’ = \(\frac{d u}{d x}\) and v’= \(\frac{d v}{d x}\) = (-sint)
g'(t) = uv’ + vu’
g'(t) = t 3 (-sin t) + cos t (3t 2 )
= -t 3 sin t + 3t 2 cos t

y = e x sin x
⇒ y = uv’ + vu’
Now u = e x ⇒ u’ = \(\frac{d u}{d x}\) e x
v = sin x ⇒ v’ = \(\frac{d v}{d x}\) cos x
i.e. y’ = e x (cos x) + sin x (e x )
= e x [sin x + cos x]

y = tan θ (sin θ + cos θ)
\(\frac{d y}{d x}\) = tan θ (cos θ – sin θ) + (sin θ + cos 0) sec 2 θ
= tan θ cos θ – tan θ sin θ + sin θ sec 2 θ + cos θ sec 2 θ
= sin θ – sin 2 θ sec θ + tan θ sec θ + sec θ
= sin θ + (1 – sin 2 θ) sec θ + sec θ tan θ
= sin θ + cos 2 θ × \(\frac{1}{\cos \theta}\) + sec θ tan θ
= sin θ + cos θ + sec θ tan θ

y = cosec x. cot x
\(\frac{d y}{d x}\) = cosec x × – cosec 2 x + cot x × – cosec x cot x
= – cosec 3 x – cosec x cot 2 x
= – cosec x (cosec 2 x + cot 2 x)

Let = u = x 2 + 4x + 6
⇒ \(\frac{d u}{d x}\) = 2x + 4
Now y = u 5 ⇒ \(\frac{d y}{d x}\) = 5u 4
∴ \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\) = 5u 4 (2x + 4)
= 5(x 2 + 4x + 6) 4 (2x + 4)
= 5 (2x + 4) (x 2 + 4x + 6) 4

y = tan 3x
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sec 2 3x. \(\frac{\mathrm{d}}{\mathrm{d} x}\) (3x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sec 2 3x × 3
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 3 sec 2 3x

Put u = tan x
\(\frac{d u}{d x}\) = sec 2 x
Now y = cos u ⇒ \(\frac{d u}{d x}\) = -sin u
Now \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\)
= (-sin u) (sec 2 x)
= -sec 2 (sin (tan x))

y = sin (e x )
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
y = cos (e x ). \(\frac{\mathrm{d}}{\mathrm{d} x}\) (e x )
y = cos ((e x )). e x
y = e x cos (e x )

F(x) = (x 3 + 4x) 7
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
F’ (x) = 7 (x 3 + 4x) 7 – 1 \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x 3 + 4x)
= 7 (x 3 + 4x) 6 (3x 2 + 4)
= 7 (3x 2 + 4) (x 3 + 4x) 6

y = cos (a 3 + x 3 )
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – sin (a 3 + x 3 ) \(\frac{\mathrm{d}}{\mathrm{d} x}\) (a 3 + x 3 )
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – sin (a 3 + x 3 ) (0 + 3x 2 )
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – 3x 2 sin (a 3 + x 3 )

y = e -mx
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = e -mx × \(\frac{\mathrm{d}}{\mathrm{d} x}\) (- mx)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = e -mx × – m
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – m e -mx = – my

y = 4 sec 5x
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 4 × sec 5x × tan 5x \(\frac{\mathrm{d}}{\mathrm{d} x}\) (5x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 4 sec 5x tan 5x × 5 × 1
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 20 sec 5x tan 5x

y = xe -x2
y = uv where u = x and v = e -x2
Now u’ = 1 and v’ = e -x2 (-2x)
v’ = – 2xe -x2
Now y = uv ⇒ y’ = uv’ + vu’
(i.e.) \(\frac{d y}{d x}\) = x[-2xe -x2 ] + e -x2 (1)
= e -x2 (1 – 2x 2 )

y = tan (cos x)
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sec 2 (cos x) × \(\frac{\mathrm{d}}{\mathrm{d} x}\) (cos x)
= sec 2 (cos x) × – sin x
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = -sin x. sec 2 (cos x)

y = \(\frac{\sin ^{2} x}{\cos x}\)
= sin x (2 + tan 2 x)
= sin x (1 + 1 + tan 2 x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sin x (1 + sec 2 x)

y = sin 3 x + cos 3 x
Here u = sin 3 x = (sin x) 3
⇒ \(\frac{d u}{d x}\) = 3 (sin x) 2 (cos x)
= 3sin 2 x cos x
v = cos 3 x = (cos x) 3
⇒ \(\frac{d v}{d x}\) = 3 (cos x) 2 (-sin x) = -3 sin x cos 2 x
Now y = u + v ⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
= 3 sin 2 x cos x – 3sin x cos 2 x
= 3 sin x cos x (sin x – cos x)

y = sin 2 (cos kx)
y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 2 sin (cos kx) × cos (cos kx) × -sin kx × k × 1
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sin (2 cos kx) × -k sin kx
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = -k sin kx. sin (2 cos kx)

y = (1 + cos 2 ) 6
y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 6(1 + cos 2 x) 6-1 (0 + 2 cos x × -sin x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 6(1 + cos 2 x) 5 × – 2 sin x cos x
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = -6 sin 2x (1 + cos 2 ) 5

y = e x cos x
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = e x cos x (x – sinx + cos x. 1)
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = e x cos x (cos x – x sin x)

y = x cos x
Taking log on both sides
log y = log x cos x
log y = cos x log x
Differentiating with respect to x

y = x log x + (log x) x
Let u = x log x, v = (log x) x
log u = log x log x
log u = (log x) (log x)
log u = (log x) 2
v = (log x) x
log v = log (log x) x
log v = x log (log x)

x y = y x
Taking log on both sides
log x y = log y x
y log x = x log y
Differentiate with respect to x

y = (cos x) log x
Taking log on both sides
log y = log (cos x) log x
log y = (log x) log (cos x)
Differentiating with respect to x

x = a (cos t + t sin t), y = a (sin t – t cos t)
\(\frac{d x}{d t}\) = a [- sin t + t cos t + sin t ]
\(\frac{d x}{d t}\) = at cos t —— (1)
y = a (sin t – t cos t)
\(\frac{d x}{d t}\) = a [cos t – (t × – sin t + cos t × 1)]
\(\frac{d x}{d t}\) = a[cos t + t sin t – cos t]
\(\frac{d x}{d t}\) = at sin t —— (2)
From equations (1) and (2) we get

Let y = sin -1 (3x – 4x 3 )
Put x = sin θ
y = sin -1 (3 sin θ – 4 sin 3 θ)
y = sin -1 (sin 3θ)
y = 3θ
y = 3 sin -1 x
\(\frac{d y}{d x}=\frac{3}{\sqrt{1-x^{2}}}\)

Let u = sin x 2
\(\frac{\mathrm{d} \mathrm{u}}{\mathrm{d} x}\) = cos (x 2 ) × 2x
\(\frac{\mathrm{d} \mathrm{u}}{\mathrm{d} x}\) = 2x cos (x 2 )
Let v = x 2

y = e tan-1x
y = e tan-1x \(\left(\frac{1}{1+x^{2}}\right)\)
⇒ y’ = \(\frac{y}{1+x^{2}}\) ⇒ y'(1 + x 2 ) = y
differentiating w.r.to x
y’ (2x) + (1 + x 2 ) (y”) = y’
(i.e.) (1 + x 2 ) y” + y’ (2x) – y’ = 0
(i.e.) (1 + x 2 ) y” + (2x – 1) y’ = 0

Given sin y = x sin (a + y) ——- (1)
Differentiating with respect to x, we get
cos y \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = x cos (a + y) (0 + \(\frac{\mathrm{dy}}{\mathrm{d} x}\)) + sin (a + y). 1

(4) 10
Explaination:
y = f(x 2 + 2)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f’ (x 2 + 2) × 2x
\(\frac{\mathrm{dy}}{\mathrm{d} x} / x=1\) = f’ (1 2 + 2) × 2 × 1
= f’(3) × 2
= 5 × 2 = 10

(3) both irrational
Explaination:
f(x) = x 2 – 3x
f’(x) = 2x – 3
f(x) = f'(x)
⇒ x 2 – 3x = 2x – 3
x 2 – 3x – 2x + 3 = 0
x 2 – 5x + 3 = 0

(3) 3
Explaination:
y = mx+c
\(\frac{d y}{d x}\) = m
y = x + c (i.e.) f(x) = x + c
y(a tx = 0) = f(0) 0 + c = 1 ⇒ c = 1
y = x + 1 ⇒ f(x) = x + 1
f(2) = 2 + 1 = 3

(4) 2
Explaination:
y = (ax – 5)e 3x
\(\frac{d y}{d x}\) = y’ = (ax – 5) (3e 3x ) + e 3x (a)
= e 3x [3ax – 15 + a]
Given \(\frac{d y}{d x}\) = -13 at x = 0
⇒ [-15 + a] = -13
⇒ a = -13 + 15
a = 2

(2) – 1
Explaination:
From equations (1) and (2) we get
f’ (3 – ) ≠ f’ (3 + )
∴ limit of f(x) does not exist at x = 3
f’ (x) does not exist at x = 3

(1) f(x) is not differentiable at x = a
Explaination:
f’ (a + ) = 3 ………. (2)
From equations (1) and (2) we get
f'(a – ) ≠ f'(a + )
∴ f’ (x) does not exist at x = a
∴ f(x) is not differentiable at x = a

(2) 2
Explaination:
f(x) = |x – 1| + |x – 3| + sin x is not differentiable at x = 1, and x = 3