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.
Let G(t) = t²½, then h(x) = ∫x³-12 G(t) dt.
Using Chain Rule, we have:
h'(x) = G(x³-12) * d/dx (x³-12)
Applying Power Rule, we get:
h'(x) = (x³-12)²½ * 3x²
Simplifying the expression:
h'(x) = 3x²√(x³-12)
Therefore, h'(x) = 3x²√(x³-12) is the derivative of h(x) for x ≥ 0.
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