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 $[d(w+ce^w)]/(dw)$ ,

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

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

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

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

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

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

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