How do you differentiate #z=w^(3/2)(w+ce^w)#?

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Answer 1 · 2017-04-28T14:18:42

#color(red)( dz/(dw) = sqrt(w)/2 [5w+ce^w(2w+3)])#

#z = w^(3/2)(w+ce^w)#

Assuming that c is a constant,

#dz/(dw) = [d[w^(3/2)(w+ce^w)]]/(dw)#

#=> dz/(dw) = w^(3/2)*[d(w+ce^w)]/(dw) + (w+ce^w)[d(w^(3/2))]/(dw)#

Using sum rule to evaluate #[d(w+ce^w)]/(dw)#,

#=> dz/(dw) = w^(3/2)*[(dw)/(dw) + c(de^w)/(dw)] + (w+ce^w)(3/2w^(3/2-1))#

#=> dz/(dw) = w^(3/2)(1+ce^w) + 3/2(w+ce^w)w^(1/2)#

#=> dz/(dw) = w ^(1/2)[w(1+ce^w) + 3/2w + 3/2ce^w]#

#=> dz/(dw) = sqrt(w)[w+ c*we^w + 3/2w + 3/2ce^w]#

#=> dz/(dw) = sqrt(w)[5/2w + 1/2ce^w(2w+3)]#

#=>color(red)( dz/(dw) = sqrt(w)/2 [5w+ce^w(2w+3)])#