For $-4 \leq x \leq 4$ , Define $F(x)=\int_{-4}^{x} \sqrt{16-t^{2}} d t$

question 72

Multiple Choice

For $-4 \leq x \leq 4$ , define $F(x) =\int_{-4}^{x} \sqrt{16-t^{2}} d t$ .Find F'(x) .

• A) \(\sqrt{2 x}\)
• B) \(-\sqrt{2 x}\)
• C) \(-\sqrt{16-x^{2}}\)
• D) \(\sqrt{16-x^{2}}\)

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