Determine a function $h$ $h$ such that $\frac{d}{d x} [\sqrt{h (x)}] = \frac{3 x^{2}}{2 \sqrt{x^{3} + 4}}$ $d/(dx) [sqrt(h(x))]=(3x^2)/(2sqrt(x^3+4))$ ?

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Answer 1 · 2017-03-18T21:25:02

$h (x) = \sqrt{{(\sqrt{x^{3} + 4} + C)}^{2}}$ $h(x) = sqrt( (sqrt( x^3 + 4) + C )^2 )$

If:

$\frac{d}{d x} [\sqrt{h (x)}] = \frac{3 x^{2}}{2 \sqrt{x^{3} + 4}}$ $d/(dx) [sqrt(h(x))]=(3x^2)/(2sqrt(x^3+4))$

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

Determine a function hhh such that d/(dx) [sqrt(h(x))]=(3x^2)/(2sqrt(x^3+4))ddx[√h(x)]=3x22√x3+4d/(dx) [sqrt(h(x))]=(3x^2)/(2sqrt(x^3+4))?