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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 42 of 72
Marks: +1, -0
Find the derivative of cos x from first principle.
Solution:  
Let f(x) = cos 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→0cos⁡(x+h)−cos⁡xh\lim\limits_{h\to 0}\frac{\cos(x+h)-\cos x}{h}
=
lim⁡h→0[−2sin⁡(x+h+x2)⋅sin⁡(x+h−x2)h]\lim\limits_{h\to 0}\left[\frac{-2\sin\left(\frac{x+h+x}{2}\right)\cdot\sin\left(\frac{x+h-x}{2}\right)}{h}\right]
= lim⁡h→0[−2sin⁡(2x+h2)⋅sin⁡(h2)h]\lim\limits_{h\to 0}\left[\frac{-2\sin\left(\frac{2x+h}{2}\right)\cdot\sin\left(\frac{h}{2}\right)}{h}\right]
= lim⁡h→0[−2sin⁡(x+h2)⋅sin⁡(h2)h]\lim\limits_{h\to 0}\left[\frac{-2\sin\left(x+\frac{h}{2}\right)\cdot\sin\left(\frac{h}{2}\right)}{h}\right]
= lim⁡h→0[−sin⁡(x+h2)⋅sin⁡(h2)h2]\lim\limits_{h\to 0}\left[\frac{-\sin\left(x+\frac{h}{2}\right)\cdot\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]
= lim⁡h→0[−sin⁡(x+h2)⋅sin⁡(h2)h2]\lim\limits_{h\to 0}\left[-\sin\left(x+\frac{h}{2}\right)\cdot\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]
= - [lim⁡h→0sin⁡(x+h2)][lim⁡h→0sin⁡(h2)h2]\left[\lim\limits_{h\to 0}\sin\left(x+\frac{h}{2}\right)\right]\left[\lim\limits_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right] = - sin x
∴ ddx\frac{d}{dx} (cos x) = - sin x.
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