Class 12 Maths Continuity and Differentiability NCERT Solutions Exercise 5.4
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Differentiate the following w.r.t. x
Question 1 exsinx Solution
Let y=exsinx
Using quotient rule, we have dydx=sinx.ddx(ex)−ex.ddx(sinx)sin2x =sinx.(ex)−ex.(cosx)sin2x =ex(sinx−cosx)sin2x,x≠nπ,n∈Z
Question 2 esin−1x Solution
Let y=y=esin−1x
Using chain rule, we have dydx=ddx(esin−1x) dydx=esin−1x.ddx(sin−1x) =esin−1x.1√1−x2 =esin−1x√1−x2 dydx=esin−1x√1−x2,x∈(−1,1)
Question 3 ex3 Solution
Let y=ex3
Using chain rule, we have dydx=ddx(ex3)=ex3.3x2=3x2.ex3
Question 4 sin(tan−1e−x) Solution
Let y=sin(tan−1e−x)
Using chain rule, we have dydx=ddx[sin(tan−1e−x)] =cos(tan−1e−x).ddx(tan−1e−x) =cos(tan−1e−x).11+(e−x)2(tan−1e−x).ddx(e−x) =cos(tan−1e−x)1+e−2x.e−x.ddx(−x) =−cos(tan−1e−x)1+e−2x.e−x
Question 5 log(cosex) Solution
Let y=log(cosex)
Using chain rule, we have dydx=ddx[log(cosex)] =1cosex.ddx(cosex) =1cosex.(−sinex)ddx(ex) =−sinexcosex.(ex) =−tanex.(ex)
Question 6 ex+ex2+ex3+ex4+ex5 Solution
Let y=ex+ex2+ex3+ex4+ex5
Using chain rule, we have dydx=ddx[ex+ex2+ex3+ex4+ex5] =ddxex+ddxex2+ddxex3+ddxex4+ddxex5 =ex+2xex2+3x2ex3+4x3ex4+5x4ex5
Question 7 √e√x Solution
Let y=√e√x
Then y2=e√x
Using chain rule, we have 2ydydx=ddx[e√x] 2ydydx=e√xddx√x 2ydydx=e√x12√x dydx=e√x14y√x dydx=e√x4√xe√x
Question 8 log(logx),x>1 Solution
Let y=log(logx)
Using chain rule, we have dydx=1logxddxlogx =1logx1x =1xlogx