Example : 44 x x2
Find f′(x) if f(x) =
x3
1 2x 3 x 2 0 2 6x
Solution
f′(x) =
d d 2 (x) (x ) dx dx 1 2x 0 2
1 2x 3 x 2
=
1 2x 3 x 2 0 2 6x
+
x x2
=0+0+
d 3 (x ) dx + 3x 2 6x
x x2
x3
0 0
6x 6x
2 2
d d (1) (2x ) dx dx 0 2
x x2
+
x2
x
x3 d (3 x 2 ) + dx 6x
x
x2
1 2x d d (0 ) ( 2) dx dx
x3 3x 2 d (6 x ) dx
x3
1 2x 3 x 2 0 0 6
x3
1 2x 3 x 2 0 0 6
= 6 (2x2 – x2) = 6x2
Example : 45 Differentiate y = cos–1
1− x2 1+ x2
with respect to z = tan–1x. Also discuss the differentiability of this function.
Solution The given function is y = cos–1
1− x2 1+ x2
Substitute x = tan θ 1 − tan 2 θ
⇒
y = cos–1
⇒
y = 2θ = 2tan–1 x
1 + tan 2 θ
= cos–1 (cos 2θ)
for ⇒ ⇒ and y = – 2θ = – 2 tan–1x for ⇒ ⇒ So the given function reduces to :
0 ≤ 2θ ≤ π 0 ≤ θ ≤ π/2 0≤x<∞ –π < 2θ < 0 –π/2 < θ < 0 –∞ < x < 0
⎧⎪ 2 tan −1 x , x ≥ 0 y= ⎨ ⎪⎩− 2 tan −1 x , x < 0 Differentiating with respect to tan–1x, we get ⎧ 2 x≥0 = ⎨ d(tan x ) ⎩− 2 x < 0
dy
−1
Alternate Method y = cos–1
1− x2 1+ x2
Differentiating with respect to x, we get
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