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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 36 of 72
Marks: +1, -0
Find the derivative of the following functions from first principle.
(i) x3x^3 – 27
(ii) (x – 1)(x – 2)
(iii) 1x2\frac{1}{x^2}
(iv) x+1x−1\frac{x+1}{x-1}
Solution:  
(i) Let f (x) = x3x^3 – 27
We have, ddx\frac{d}{dx} (f (x)) = lim⁡h→0f(x+h)−f(x)h\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}
= lim⁡h→0[(x+h)3−27]−[x3−27]h\lim\limits_{h\to 0}\frac{[(x+h)^3-27]-[x^3-27]}{h}
=
lim⁡h→0[x3+3x2h+3xh2+h3−27−x3+27h]\lim\limits_{h\to 0}\left[\frac{x^3+3x^2h+3xh^2+h^3-27-x^3+27}{h}\right]
= lim⁡h→03x2h+3xh2+h3h\lim\limits_{h\to 0}\frac{3x^2h+3xh^2+h^3}{h}
= lim⁡h→0[3x2+3xh+h2h]\lim\limits_{h\to 0}\left[\frac{3x^2+3xh+h^2}{h}\right]
= lim⁡h→0(3x2+3xh+h2)\lim\limits_{h\to 0}(3x^2+3xh+h^2)
= 3x23x^2 + 3x (0) + (0)2(0)^2 = 3x23x^2.
(ii) Let f(x) = (x – 1) (x – 2)
We have, ddx\frac{d}{dx} (f (x)) = lim⁡h→0f(x+h)−f(x)h\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}
=
lim⁡h→0(x+h−1)(x+h−2)−(x−1)(x−2)h\lim\limits_{h\to 0}\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}
=
lim⁡h→0[(x+h)2−2(x+h)−(x+h)+2]−[x2−2x−x+2]h\lim\limits_{h\to 0}\frac{[(x+h)^2-2(x+h)-(x+h)+2]-[x^2-2x-x+2]}{h}
=
lim⁡h→0[x2+h2+2xh−3(x+h)+2]−[x2−3x+2]h\lim\limits_{h\to 0}\frac{[x^2+h^2+2xh-3(x+h)+2]-[x^2-3x+2]}{h}
= lim⁡h→0h2+2xh−3hh\lim\limits_{h\to 0}\frac{h^2+2xh-3h}{h}
= lim⁡h→0\lim\limits_{h\to 0} (h + 2x - 3) = 2x - 3
(iii) Let f (x) = 1x2\frac{1}{x^2}
We have, ddx\frac{d}{dx} (f (x)) = lim⁡h→0\lim\limits_{h\to 0} f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}
= lim⁡h→0[1(x+h)2−1x2h]\lim\limits_{h\to 0}\left[\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\right] = lim⁡h→0[x2−(x+h)2(x+h)2x2h]\lim\limits_{h\to 0}\left[\frac{x^2-(x+h)^2}{(x+h)^2 x^2 h}\right] = lim⁡h→0[x2−x2−h2−2xh(x+h)2⋅x2⋅h]\lim\limits_{h\to 0}\left[\frac{x^2-x^2-h^2-2xh}{(x+h)2 \cdot x^2 \cdot h}\right]
= lim⁡h→0[−h2−2xh(x+h)2⋅x2⋅h]\lim\limits_{h\to 0}\left[\frac{-h^2-2xh}{(x+h)^2 \cdot x^2 \cdot h}\right] = lim⁡h→0(−hh)[h+2x(x+h)2⋅x2]\lim\limits_{h\to 0}\left(-\frac{h}{h}\right)\left[\frac{h+2x}{(x+h)^2 \cdot x^2}\right]
= - lim⁡h→0(h+2x(x+h)2⋅x2)\lim\limits_{h\to 0}\left(\frac{h+2x}{(x+h)^2 \cdot x^2}\right) = −2xx2⋅x2\frac{-2x}{x^2 \cdot x^2} = −2x3\frac{-2}{x^3}
(iv) Let f (x) = x+1x−1\frac{x+1}{x-1}
We have, ddx\frac{d}{dx} (f (x)) = lim⁡h→0f(x+h)−f(x)h\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}
ddx\frac{d}{dx} (f (x)) = lim⁡h→0[x+h+1x+h−1−x+1x−1h]\lim\limits_{h\to 0}\left[\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}\right]
=
lim⁡h→0[(x+h+1)(x−1)−(x+1)(x+h−1)h(x+h−1)(x−1)]\lim\limits_{h\to 0}\left[\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{h(x+h-1)(x-1)}\right]
=
lim⁡h→0[x2+hx+x−x−h−1−[x2+hx−x+x+h−1]h(x+h−1)(x−1)]\lim\limits_{h\to 0}\left[\frac{x^2+hx+x-x-h-1-[x^2+hx-x+x+h-1]}{h(x+h-1)(x-1)}\right]
=
lim⁡h→0[(x2+hx−h−1)−(x2+hx+h−1)h(x+h−1)(x−1)]\lim\limits_{h\to 0}\left[\frac{(x^2+hx-h-1)-(x^2+hx+h-1)}{h(x+h-1)(x-1)}\right]
=
lim⁡h→0[x2+hx−h−1−x2−hx−h−1h(x+h−1)(x−1)]\lim\limits_{h\to 0}\left[\frac{x^2+hx-h-1-x^2-hx-h-1}{h(x+h-1)(x-1)}\right]
= lim⁡h→0[−2hh(x+h−1)(x−1)]\lim\limits_{h\to 0}\left[\frac{-2h}{h(x+h-1)(x-1)}\right]
= lim⁡h→0[−2(x+h−1)(x−1)]\lim\limits_{h\to 0}\left[\frac{-2}{(x+h-1)(x-1)}\right]
= −2(x+0−1)(x−1)\frac{-2}{(x+0-1)(x-1)}
= −2(x−1)2\frac{-2}{(x-1)^2}
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