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
