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Class 12 Maths Continuity and Differentiability NCERT Solutions Exercise 5.4




In this page we have Class 12 Maths Continuity and Differentiability NCERT Solutions Exercise 5.4 . Hope you like them and do not forget to like , social share and comment at the end of the page.

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(sinxcosx)sin2x,xnπ,nZ  

Question 2
esin1x
Solution

Let y=y=esin1x
Using chain rule, we have
dydx=ddx(esin1x)
dydx=esin1x.ddx(sin1x)
=esin1x.11x2
=esin1x1x2
dydx=esin1x1x2,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(tan1ex)
Solution
Let y=sin(tan1ex)
Using chain rule, we have
dydx=ddx[sin(tan1ex)]
=cos(tan1ex).ddx(tan1ex)
=cos(tan1ex).11+(ex)2(tan1ex).ddx(ex)
=cos(tan1ex)1+e2x.ex.ddx(x)
=cos(tan1ex)1+e2x.ex

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
ex
Solution
Let y=ex
Then y2=ex Using chain rule, we have
2ydydx=ddx[ex]
2ydydx=exddxx
2ydydx=ex12x
dydx=ex14yx
dydx=ex4xex

Question 8
log(logx),x>1
Solution
Let y=log(logx)
Using chain rule, we have
dydx=1logxddxlogx
=1logx1x
=1xlogx

 


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