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Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.4 | Set 2

  • Last Updated : 16 May, 2021
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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,

\frac{dy}{dx}=\frac{d}{dx}[(x sin x + cos x) (x cos x − sin x)]



On using product rule we get,

(xcosx−sinx)\frac{d}{dx}(xsinx+cosx)+(xsinx+cosx)\frac{d}{dx}(xcosx-sinx)

On using chain rule, we get,

(xcosx−sinx)\left[\frac{d}{dx}(xsinx)+\frac{d}{dx}(cosx)\right]+(xsinx+cosx)\left[\frac{d}{dx}(xcosx)-\frac{d}{dx}(sinx)\right]

On using product rule again, we get,

(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]

(xcosx−sinx)\left[sinx+xcosx-sinx\right]+(xsinx+cosx)\left[cosx-xsinx-cosx\right]

= (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,

\frac{dy}{dx}=\frac{d}{dx}[(x sin x + cos x) (e^x + x^2 log x)]

On using product rule we get,



(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)

On using chain rule, we get,

(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]

On using product rule again, we get,

(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]

(e^x+x^2log x)\left[sinx+xcosx-sinx\right]+(xsinx+cosx)\left[e^x+2xlogx+x^2(\frac{1}{x})\right]

(e^x+x^2log x)\left(xcosx\right)+(xsinx+cosx)\left[e^x+2xlogx+x\right]

= (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,

\frac{dy}{dx}=\frac{d}{dx}[(1−2tanx)(5+4sinx)]

On using product rule we get,

(5+4sinx)\frac{d}{dx}(1−2tanx)+(1-2tanx)\frac{d}{dx}(5+4sinx)

(5+4sinx)(−2sec^2x)+(1-2tanx)(4cosx)

= −10 sec2 x − 8 sin x sec2 x + 4 cos x − 8 tan x cos x

−10sec^2x−8sinx(\frac{1}{cos^2x})+4cosx−8(\frac{sinx}{cosx})cosx

= −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,

\frac{dy}{dx}=\frac{d}{dx}[(1 + x^2) cos x]

On using product rule we get,

cosx\frac{d}{dx}[(1 + x^2)]+(1+x^2)\frac{d}{dx}(cosx)

= 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,

\frac{dy}{dx}=\frac{d}{dx}[(sin x) (sin x)]

On using product rule we get,

sinx\frac{d}{dx}(sinx)+(sinx)\frac{d}{dx}(sinx)

= sin x cos x + sin x cos x

= 2 sin x cos x

= sin 2x

Question 16. Differentiate log_{x^2}x with respect to x.

Solution:



We have, 

=> y = log_{x^2}x

\frac{logx}{logx^2}

\frac{logx}{2logx}

\frac{1}{2}

On differentiating both sides, we get,

\frac{dy}{dx}=\frac{d}{dx}(\frac{1}{2})

= 0

Question 17. Differentiate e^xlog\sqrt{x}tanx with respect to x.

Solution:

We have, 

=> y = e^xlog\sqrt{x}tanx

On differentiating both sides, we get,

\frac{dy}{dx}=\frac{d}{dx}(e^xlog\sqrt{x}tanx)

On using product rule we get,

log\sqrt{x}tanx\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(log\sqrt{x}tanx)

On using product rule again, we get,

log\sqrt{x}tanx(e^x)+e^x\left[tanx\frac{d}{dx}(log\sqrt{x})+log\sqrt{x}\frac{d}{dx}(tanx)\right]

e^xlog\sqrt{x}tanx+e^x\left[tanx(\frac{1}{\sqrt{x}})(\frac{1}{2\sqrt{x}})+log\sqrt{x}sec^2x\right]

e^xlog\sqrt{x}tanx+e^x\left[\frac{tanx}{2x}+log\sqrt{x}sec^2x\right]

\frac{1}{2}e^xlogxtanx+e^x\left[\frac{tanx}{2x}+\frac{1}{2}logxsec^2x\right]



\frac{e^x}{2}[logxtanx+\frac{tanx}{x}+logxsec^2x]

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,

\frac{dy}{dx}=\frac{d}{dx}(x^3e^xcosx)

On using product rule we get,

e^xcosx\frac{d}{dx}(x^3)+x^3\frac{d}{dx}(e^xcosx)

On using product rule again, we get,

e^xcosx(3x^2)+x^3[cosx\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(cosx)]

3x^2e^xcosx+x^3[e^xcosx-e^xsinx]

3x^2e^xcosx+x^3e^xcosx-x^3e^xsinx

x^2e^x(3cosx+xcosx-xsinx)

Question 19. Differentiate \frac{x^2cos\frac{\pi}{4}}{sinx} with respect to x.

Solution:

We have, 

=> y = \frac{x^2cos\frac{\pi}{4}}{sinx}

=> y = x^2cos\frac{\pi}{4}cosecx

On differentiating both sides, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^2cos\frac{\pi}{4}cosecx)

On using product rule we get,



cos\frac{\pi}{4}cosecx\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(cos\frac{\pi}{4}cosecx)

On using product rule again, we get,

cos\frac{\pi}{4}cosecx(2x)+x^2[cosecx\frac{d}{dx}(cos\frac{\pi}{4})+cos\frac{\pi}{4}\frac{d}{dx}(cosecx)]

2xcos\frac{\pi}{4}cosecx+x^2[cosecx(0)+cos\frac{\pi}{4}(-cosecxcotx)]

2xcos\frac{\pi}{4}cosecx+x^2[cos\frac{\pi}{4}(-cosecxcotx)]

2xcos\frac{\pi}{4}cosecx-x^2cos\frac{\pi}{4}cosecxcotx

2x(\frac{1}{\sqrt{2}})cosecx-x^2(\frac{1}{\sqrt{2}})cosecxcotx

\frac{xcosecx}{\sqrt{2}}(2-xcotx)

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,

\frac{dy}{dx}=\frac{d}{dx}[x^4(5sinx−3cosx)]

On using product rule we get,

(5sinx−3cosx)\frac{d}{dx}(x^4)+x^4\frac{d}{dx}(5sinx−3cosx)

(5sinx−3cosx)(4x^3)+x^4(5cosx+3sinx)

= 20 x3 sin x − 12 x3 cos x + 5x4 cos x + 3x4 sin x

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