To find [tex]\( F'(x) \)[/tex] when [tex]\( F(x) = \int_{x^2 - 1}^0 f(t) \, dt \)[/tex]:
1. Start by applying the Leibniz rule for differentiation under the integral sign. The general form of this rule is:
[tex]\[
\frac{d}{dx} \left( \int_{u(x)}^{v(x)} f(t) \, dt \right) = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)
\][/tex]
where [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex] are the bounds of integration.
2. In our case, [tex]\( v(x) = 0 \)[/tex] and [tex]\( u(x) = x^2 - 1 \)[/tex]. We need to find [tex]\( \frac{d}{dx} \left( \int_{x^2 - 1}^0 f(t) \, dt \right) \)[/tex].
3. Since [tex]\( v(x) = 0 \)[/tex] is a constant, [tex]\( v'(x) = 0 \)[/tex]. So the term involving [tex]\( v(x) \)[/tex] will be zero:
[tex]\[
f(0) \cdot 0 = 0
\][/tex]
4. Now consider the lower bound [tex]\( u(x) = x^2 - 1 \)[/tex]. Its derivative with respect to [tex]\( x \)[/tex] is:
[tex]\[
u'(x) = \frac{d}{dx}(x^2 - 1) = 2x
\][/tex]
5. Therefore, applying the Leibniz rule, [tex]\( F'(x) \)[/tex] becomes:
[tex]\[
F'(x) = -f(u(x)) \cdot u'(x) = -f(x^2 - 1) \cdot 2x
\][/tex]
6. Now, we need to find [tex]\( F'(3) \)[/tex]:
[tex]\[
F'(3) = -f(3^2 - 1) \cdot 2 \cdot 3 = -f(8) \cdot 6
\][/tex]
7. Given that [tex]\( f(8) = -4 \)[/tex]:
[tex]\[
F'(3) = -(-4) \cdot 6 = 4 \cdot 6 = 24
\][/tex]
Hence, [tex]\( F'(3) = 24 \)[/tex].
