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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 44 of 72
Marks: +1, -0
Find the derivative of the following functions from first principle:
(i) –x
(ii) (−x)−1(-x)^{-1}
(iii) sin (x + 1)
(iv) cos (x−π8)\left(x-\frac{\pi}{8}\right)
Solution:  
(i) Let f(x) = – x
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)−(−x)h\lim\limits_{h\to 0}\frac{-(x+h)-(-x)}{h}
= lim⁡h→0−x−h+xh\lim\limits_{h\to 0}\frac{-x-h+x}{h}
= lim⁡h→0−hh\lim\limits_{h\to 0}\frac{-h}{h}
= lim⁡h→0(−1)\lim\limits_{h\to 0}(-1) = - 1
(ii) Let f(x) = (−x)−1(-x)^{-1}
⇒ f (x) = −1x-\frac{1}{x}
We have, ddx\frac{d}{dx} (f (x)) = lim⁡h→0[f(x+h)−f(x)h]\lim\limits_{h\to 0}\left[\frac{f(x+h)-f(x)}{h}\right]
= lim⁡h→0[−1x+h−(−1x)]h\lim\limits_{h\to 0}\frac{\left[-\frac{1}{x+h}-\left(-\frac{1}{x}\right)\right]}{h} = lim⁡h→0[−1x+h+1x]h\lim\limits_{h\to 0}\frac{\left[-\frac{1}{x+h}+\frac{1}{x}\right]}{h}
= lim⁡h→0[−x+x+hx(x+h)h]\lim\limits_{h\to 0}\left[\frac{-x+x+h}{x(x+h)h}\right]
= lim⁡h→0[hhx(x+h)]\lim\limits_{h\to 0}\left[\frac{h}{hx(x+h)}\right] = 1x2\frac{1}{x^{2}}
(iii) Let f(x) = sin (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}
= lim⁡h→0sin⁡(x+h+x)−sin⁡(x+1)h\lim\limits_{h\to 0}\frac{\sin(x+h+x)-\sin(x+1)}{h}
=
lim⁡h→02cos⁡(x+h+1+x+12)sin⁡(x+h+1−x−12)h\lim\limits_{h\to 0}\frac{2\cos\left(\frac{x+h+1+x+1}{2}\right)\sin\left(\frac{x+h+1-x-1}{2}\right)}{h}
=
lim⁡h→02[cos⁡(2(x+1)+h2)sin⁡(h2)]2×h2\lim\limits_{h\to 0}\frac{2\left[\cos\left(\frac{2(x+1)+h}{2}\right)\sin\left(\frac{h}{2}\right)\right]}{2\times\frac{h}{2}}
lim⁡h→0cos⁡(x+1+h2)(sin⁡(h2)h2)\lim\limits_{h\to 0}\cos\left(x+1+\frac{h}{2}\right)\left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)
=
[lim⁡h→0cos⁡(x+1+h2)][lim⁡h→0sin⁡(h2)h2]\left[\lim\limits_{h\to 0}\cos\left(x+1+\frac{h}{2}\right)\right]\left[\lim\limits_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]
= cos (x + 1) × (1) = cos (x + 1).
(iv) Let f (x) = cos (x−π8)\left(x-\frac{\pi}{8}\right)
We have, ddx\frac{d}{dx} (f (x)) =
lim⁡h→0cos⁡(x+h−π8)−cos⁡(x−π8)h\lim\limits_{h\to 0}\frac{\cos\left(x+h-\frac{\pi}{8}\right)-\cos\left(x-\frac{\pi}{8}\right)}{h}
=
lim⁡h→0[−2sin⁡(x+h−π8+x−π82)sin⁡(s+h−π8+x−π82)h]\lim\limits_{h\to 0}\left[\frac{-2\sin\left(\frac{x+h-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}\right)\sin\left(\frac{s+h-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}\right)}{h}\right]
=
lim⁡h→0[−2sin⁡(x−π8+h2)⋅sin⁡(h2)h]\lim\limits_{h\to 0}\left[\frac{-2\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right)\cdot\sin\left(\frac{h}{2}\right)}{h}\right]
=
[lim⁡h→0sin⁡(x−π8+h2)][lim⁡h→0sin⁡(h2)h2]\left[\lim\limits_{h\to 0}\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right)\right]\left[\lim\limits_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]
= - sin (x−π8)\left(x-\frac{\pi}{8}\right) (1)
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