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diffrentiate with respect to x .... (xcosx)^1/x+(xsinx)^x

suyogya jain , 13 Years ago
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saiomkar kandukuri

question is(xcosx)^1/x +(xsinx)^x 

take  (xcosx)^1/x =m

logm =1/x *(log(xcosx)

differentiate wrt to X

1/m  dm/dx =-1/x^2 *(log(xcosx)) + (1/x*xcosx)*(cosx-xsinx)

solve and get dm/dx

similarly take (xsinx)^x =n

logn = x(log(xsinx))

differentiate wrt to X

1/n *(dn/dx) = log(xsinx) + x*(1/Xsinx)*(sinx+xsinx)

get dn/dx from above one

dm/dx +dn/dx gives the differentiation of given question.......

 

formulaes used are 1)differentiationof logy is  1/y *(dy/dx)

2)differentiation of xsinx is sinx +xcosx

3)differentiation of xcosx is cosx -xsinx

Last Activity: 13 Years ago
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