If h(x)=∫x3−12+t2−−−−−√ⅆt for x≥0, then h′(x)=
3×2 sqrt (2 + x^6)
To find the derivative of h(x), we need to use the Fundamental Theorem of Calculus:
h(x) = ∫x^3-12+t^(1/2) dt from 0 to x
Using the Chain Rule and the Fundamental Theorem of Calculus, we obtain:
h'(x) = [x^3-12+x^(1/2)]’
h'(x) = 3x^2+1/2x^(-1/2)
Simplifying, we get:
h'(x) = 3x^2+1/2x^(1/2)
Therefore, h'(x) = 3x^2+1/2x^(1/2).
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