Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.4 | Set 2
Last Updated :
16 May, 2021
Question 11. Differentiate (x sin x + cos x) (x cos x − sin x) with respect to x.
Solution:
We have,
=> y = (x sin x + cos x) (x cos x − sin x)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}[(x sin x + cos x) (x cos x − sin x)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-cdb15fa10ae3ad47526c459c610de2cc_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com (xcosx−sinx)\frac{d}{dx}(xsinx+cosx)+(xsinx+cosx)\frac{d}{dx}(xcosx-sinx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-23f848237f8ea0f88b4c5cb4a9da43b7_l3.png)
On using chain rule, we get,
= ![Rendered by QuickLaTeX.com (xcosx−sinx)\left[\frac{d}{dx}(xsinx)+\frac{d}{dx}(cosx)\right]+(xsinx+cosx)\left[\frac{d}{dx}(xcosx)-\frac{d}{dx}(sinx)\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-8157c35f420374695c1cac0aed7584e5_l3.png)
On using product rule again, we get,
= ![Rendered by QuickLaTeX.com (xcosx−sinx)\left[sinx\frac{d}{dx}(x)+x\frac{d}{dx}(sinx)-sinx\right]+(xsinx+cosx)\left[cosx\frac{d}{dx}(x)+x\frac{d}{dx}(cosx)-cosx\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-77bce0beb17e17479b84267dd2a3b228_l3.png)
= ![Rendered by QuickLaTeX.com (xcosx−sinx)\left[sinx+xcosx-sinx\right]+(xsinx+cosx)\left[cosx-xsinx-cosx\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e4bd38066f40a3395a7ff2448812e46d_l3.png)
= (x cos x − sin x) (x cos x) + (x sin x + cos x) (−x sin x)
= x2 cos2 x − x cos x sin x − x2 sin2 x − x cos x sin x
= x2 (cos2 x − sin2 x) − 2x cos x sin x
= x2 cos 2x − x sin 2x
= x (x cos 2x − sin 2x)
Question 12. Differentiate (x sin x + cos x) (ex + x2 log x) with respect to x.
Solution:
We have,
=> y = (x sin x + cos x) (ex + x2 log x)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}[(x sin x + cos x) (e^x + x^2 log x)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c69514c0ac034add07511d35d9306168_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com (e^x + x^2 log x)\frac{d}{dx}(x sin x + cos x)+(x sin x + cos x)\frac{d}{dx}(e^x + x^2 log x)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-3b53e89ad423f60b2715fddd86289a7d_l3.png)
On using chain rule, we get,
= ![Rendered by QuickLaTeX.com (e^x + x^2 log x)\left[\frac{d}{dx}(x sin x)+\frac{d}{dx}(cos x)\right]+(x sin x + cos x)\left[\frac{d}{dx}(e^x) + \frac{d}{dx}(x^2 log x)\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-1b032e5cc9fefcd0001f5f13836e3278_l3.png)
On using product rule again, we get,
= ![Rendered by QuickLaTeX.com (e^x+x^2log x)\left[sinx\frac{d}{dx}(x)+x\frac{d}{dx}(sinx)-sinx\right]+(xsinx+cosx)\left[e^x+logx\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(logx)\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-6ff38c0606682b47ca95a6d7476a7c42_l3.png)
= ![Rendered by QuickLaTeX.com (e^x+x^2log x)\left[sinx+xcosx-sinx\right]+(xsinx+cosx)\left[e^x+2xlogx+x^2(\frac{1}{x})\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c1d3652c9dd36bc328a1c29cc7927552_l3.png)
= ![Rendered by QuickLaTeX.com (e^x+x^2log x)\left(xcosx\right)+(xsinx+cosx)\left[e^x+2xlogx+x\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-52ce02593d6c5034765e4cf95e2fb53c_l3.png)
= (x cos x) (ex + x2 log x) +(x sin x + cos x) (ex + 2x log x + x)
Question 13. Differentiate (1 − 2 tan x) (5 + 4 sin x) with respect to x.
Solution:
We have,
=> y = (1 − 2 tan x) (5 + 4 sin x)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}[(1−2tanx)(5+4sinx)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-b6a21174c3744ae73450e8d15323beff_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com (5+4sinx)\frac{d}{dx}(1−2tanx)+(1-2tanx)\frac{d}{dx}(5+4sinx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-6a38b2bf180072462c1921b52d7b768e_l3.png)
= ![Rendered by QuickLaTeX.com (5+4sinx)(−2sec^2x)+(1-2tanx)(4cosx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-8af9cfd150462b17f37c5801bb49ad9a_l3.png)
= −10 sec2 x − 8 sin x sec2 x + 4 cos x − 8 tan x cos x
= ![Rendered by QuickLaTeX.com −10sec^2x−8sinx(\frac{1}{cos^2x})+4cosx−8(\frac{sinx}{cosx})cosx](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-f30321945ed690339682a5295d126189_l3.png)
= −10 sec2 x − 8 tan x sec x + 4 cos x − 8 sin x
Question 14. Differentiate (1 + x2) cos x with respect to x.
Solution:
We have,
=> y = (1 + x2) cos x
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}[(1 + x^2) cos x]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-62913f6fd76b51875cf53f35612ab25b_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com cosx\frac{d}{dx}[(1 + x^2)]+(1+x^2)\frac{d}{dx}(cosx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-8cd5102494d0fde820290f1b699e4fc1_l3.png)
= cos x (2x) + (1 + x2) (−sinx)
= 2x cos x − sin x(1 + x2) (sinx)
Question 15. Differentiate sin2 x with respect to x.
Solution:
We have,
=> y = sin2 x
=> y = (sin x) (sin x)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}[(sin x) (sin x)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-2fb4106cc811e6d25cedf2ccc3ef6f6c_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com sinx\frac{d}{dx}(sinx)+(sinx)\frac{d}{dx}(sinx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-03786be550cc2486db3c9ff152edccbe_l3.png)
= sin x cos x + sin x cos x
= 2 sin x cos x
= sin 2x
Question 16. Differentiate
with respect to x.
Solution:
We have,
=> y = ![Rendered by QuickLaTeX.com log_{x^2}x](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-aa1dcffd2a727b02c1335dcaf618e63c_l3.png)
= ![Rendered by QuickLaTeX.com \frac{logx}{logx^2}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-cf1b4e301bcbbd51773205f488c8d4da_l3.png)
= ![Rendered by QuickLaTeX.com \frac{logx}{2logx}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e8eb7224f9b7ef61288081919cc01673_l3.png)
= ![Rendered by QuickLaTeX.com \frac{1}{2}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-5f44a2631ffeb8ab30b75e50e1080f7e_l3.png)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}(\frac{1}{2})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-cb1e33c14a2cbde757f08cf00ebca660_l3.png)
= 0
Question 17. Differentiate
with respect to x.
Solution:
We have,
=> y = ![Rendered by QuickLaTeX.com e^xlog\sqrt{x}tanx](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-632cc29c7363cce4f1314d68cddf7af0_l3.png)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}(e^xlog\sqrt{x}tanx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-fb40acd6ca20d9685d38cc992682a616_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com log\sqrt{x}tanx\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(log\sqrt{x}tanx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-bb4db034efc12d55779f0b7e2352f459_l3.png)
On using product rule again, we get,
= ![Rendered by QuickLaTeX.com log\sqrt{x}tanx(e^x)+e^x\left[tanx\frac{d}{dx}(log\sqrt{x})+log\sqrt{x}\frac{d}{dx}(tanx)\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d3f0ad520d94a3327fd4513a61af2795_l3.png)
= ![Rendered by QuickLaTeX.com e^xlog\sqrt{x}tanx+e^x\left[tanx(\frac{1}{\sqrt{x}})(\frac{1}{2\sqrt{x}})+log\sqrt{x}sec^2x\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-1d4ba44c6188ccaba717ca7f9f7fcc5f_l3.png)
= ![Rendered by QuickLaTeX.com e^xlog\sqrt{x}tanx+e^x\left[\frac{tanx}{2x}+log\sqrt{x}sec^2x\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ce8a3de34cd9fdcf02ad2f39a0ec66ff_l3.png)
= ![Rendered by QuickLaTeX.com \frac{1}{2}e^xlogxtanx+e^x\left[\frac{tanx}{2x}+\frac{1}{2}logxsec^2x\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ea3750023d8844469c99a0f3cdbbb478_l3.png)
= ![Rendered by QuickLaTeX.com \frac{e^x}{2}[logxtanx+\frac{tanx}{x}+logxsec^2x]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c1309defe8695ed51ab93641da6f14b5_l3.png)
Question 18. Differentiate x3 ex cos x with respect to x.
Solution:
We have,
=> y = x3 ex cos x
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}(x^3e^xcosx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-18458b920298e71f04bfd53aca217084_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com e^xcosx\frac{d}{dx}(x^3)+x^3\frac{d}{dx}(e^xcosx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ce91fe3503ff3879c069ee8fb54ded31_l3.png)
On using product rule again, we get,
= ![Rendered by QuickLaTeX.com e^xcosx(3x^2)+x^3[cosx\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(cosx)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-a3d45fff4320d95df5d524bc49cb3f3d_l3.png)
= ![Rendered by QuickLaTeX.com 3x^2e^xcosx+x^3[e^xcosx-e^xsinx]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-01b70308e1c5b25a38b77ae4c25a7ee5_l3.png)
= ![Rendered by QuickLaTeX.com 3x^2e^xcosx+x^3e^xcosx-x^3e^xsinx](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-926b2c0ad81fac36c2319c4f82b8b8d2_l3.png)
= ![Rendered by QuickLaTeX.com x^2e^x(3cosx+xcosx-xsinx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d2bde9c4e1091fcc607f405d25e737e2_l3.png)
Question 19. Differentiate
with respect to x.
Solution:
We have,
=> y = ![Rendered by QuickLaTeX.com \frac{x^2cos\frac{\pi}{4}}{sinx}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-398ce95f96faae727542ea6a6947dd20_l3.png)
=> y = ![Rendered by QuickLaTeX.com x^2cos\frac{\pi}{4}cosecx](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-1258169c7cc6d5a3c5f1771292a88264_l3.png)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}(x^2cos\frac{\pi}{4}cosecx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-420ba8cf7a52040a852b29f8e8f15cb3_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com cos\frac{\pi}{4}cosecx\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(cos\frac{\pi}{4}cosecx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c901abf738c1498755783f4f877f788a_l3.png)
On using product rule again, we get,
= ![Rendered by QuickLaTeX.com cos\frac{\pi}{4}cosecx(2x)+x^2[cosecx\frac{d}{dx}(cos\frac{\pi}{4})+cos\frac{\pi}{4}\frac{d}{dx}(cosecx)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-05b6feb967aebd838fb10d3a06dd4394_l3.png)
= ![Rendered by QuickLaTeX.com 2xcos\frac{\pi}{4}cosecx+x^2[cosecx(0)+cos\frac{\pi}{4}(-cosecxcotx)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-6798cb2a637f0f1f368cbc7b0e1115e3_l3.png)
= ![Rendered by QuickLaTeX.com 2xcos\frac{\pi}{4}cosecx+x^2[cos\frac{\pi}{4}(-cosecxcotx)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-1fa727c8125af7ab5f67837ddd3ec65c_l3.png)
= ![Rendered by QuickLaTeX.com 2xcos\frac{\pi}{4}cosecx-x^2cos\frac{\pi}{4}cosecxcotx](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-dcfc03e3a6d35fea663e12efd1401990_l3.png)
= ![Rendered by QuickLaTeX.com 2x(\frac{1}{\sqrt{2}})cosecx-x^2(\frac{1}{\sqrt{2}})cosecxcotx](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ff1652ee49e47c78371516458162efdd_l3.png)
= ![Rendered by QuickLaTeX.com \frac{xcosecx}{\sqrt{2}}(2-xcotx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-49d27e4c9b930ba7cafe883d7c75d21a_l3.png)
Question 20. Differentiate x4 (5 sin x − 3 cos x) with respect to x.
Solution:
We have,
=> y = x4 (5 sin x − 3 cos x)
On differentiating both sides, we get,
![Rendered by QuickLaTeX.com \frac{dy}{dx}=\frac{d}{dx}[x^4(5sinx−3cosx)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-8e7a031127c9f080e239d29e5f9b71e3_l3.png)
On using product rule we get,
= ![Rendered by QuickLaTeX.com (5sinx−3cosx)\frac{d}{dx}(x^4)+x^4\frac{d}{dx}(5sinx−3cosx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-72ebf9ce01fa3f0a450cc656f27806ac_l3.png)
= ![Rendered by QuickLaTeX.com (5sinx−3cosx)(4x^3)+x^4(5cosx+3sinx)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d794b9b90013f5b81004f8b67bd69e0c_l3.png)
= 20 x3 sin x − 12 x3 cos x + 5x4 cos x + 3x4 sin x
Share your thoughts in the comments
Please Login to comment...