Examples
Ex1:
Let f: (–1, 1) → R be a differentiable function with f(0) = –1 and f∣(0) = 1.
Let g(x) = [f(2f(x) + 2)]2. Then find g'(0).
Sol:
Given g(x) = [f(2 f(x) + 2)]2
⇒ g'(x) = 2 f(2 f(x) + 2).f '(2 f(x) + 2).2f '(x)
Substituting x = 0,
⇒ g'(0) = 2f(2 f(0) + 2).f '(2f(0) + 2).2f '(0)
= 2f(0).f '(0).2f '(0) (∴ f(0) = –1 and f '(0) = 1)
= 2(–1)(1).2(1) = – 4
Ex2:
Find derivative of following functions ?
(i) y = ln3 tan2(x4)
(ii) exp (cos3(tan– 1x3)2)
Sol :
(i) y = ln3 (tan2(x4))
(ii) y = exp (cos3(tan x3)2)
= exp(cos3(tan x3)2). (cos3(tan x3)2)
= y.3 cos2(tan x3)2.cos(tan x3)2
= 3y cos2(tan x3)2 (– sin(tan x3)2).(tan x3)2
= – 3y cos2(tan x3)2 sin(tan x3)2) (2 tan (x3).sec2 x3).3x2
= – 18y x2 cos2(tan x3)2.sin(tan x3)2 tan(x3) sec2 x3.
Ex3:
Find .
Sol:
By differentiating both sides of the given equation with respect to 'x', we get:
Ex4:
Sol:
Ex5:
Sol:
Ex6:
If y5 + xy2 + x3 = 4x + 3, then find at (2, 1)
Sol:
y5 + xy2 + x3 = 4x + 3
differentiating w.r.t. x
Ex7:
If y = x cosy + y cosx, find
Sol:
Method 1:
Given y = x cosy + y cosx,
Differentiating both sides with respect to x, we get
Method 2:
Shortcut method : y = x cosy + y cosx
Let f = x cosy + y cosx – y
⇒ = cosy –
y sinx and = – x siny + cosx – 1
Ex8:
Sol:
Given function is y =
or y + y2 + y cosx = (1 + y)sinx
On differentiating both sides w.r.t. x we get
+ 2y + y(– sinx) + cosx. = (1 + y)cosx + .sinx
⇒ {1 + 2y + cosx – sinx} = (1 + y)cosx + y sinx
⇒
Ex9:
The equation y2exy = 9 e– 3x2 defines y as a differentiable function of x. The value of for x = –1 and y = 3
Sol: