Determine a function #h# such that #d/(dx) [sqrt(h(x))]=(3x^2)/(2sqrt(x^3+4))#?

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Answer 1 · 2017-03-18T21:25:02

#h(x) = sqrt( (sqrt( x^3 + 4) + C )^2 )#

If:

#d/(dx) [sqrt(h(x))]=(3x^2)/(2sqrt(x^3+4))#

Then:

#int \ d/(dx) [sqrt(h(x))] \ dx =int \ (3x^2)/(2sqrt(x^3+4)) \ dx#

#implies sqrt(h(x)) =int \ (3x^2)/(2sqrt(x^3+4)) \ dx#

For:

#int \ (3x^2)/(2sqrt(x^3+4)) \ dx#

We say that: #z = x^3 + 4, dz = 3x^2 dx#

#implies int \ (3x^2)/(2sqrt(z)) \ (dz)/(3x^2)#

#= 1/2 int \ (dz)/(sqrt(z)) \ #

#= sqrt z + C#

#= sqrt( x^3 + 4) + C color(red)(= sqrt(h(x))#

So

#h(x) = sqrt( (sqrt( x^3 + 4) + C )^2 )#

Answer 2 · 2017-03-18T21:32:34

#h(x) = x^3+4#

#d/dx (sqrt(h(x))) = 1/2(h(x))^(-1/2) d/dx(h(x)) = (h'(x))/(2sqrt(h(x))#

If #h(x) = x^3+4# then #h'(x) = 3x^2#

#d/dx (sqrt(x^3+4)) = 1/2(x^3+4)^(-1/2)(3x^2) = (3x^2)/(2 sqrt(x^3+4))#

So #h(x) = x^3+4#