Question

simplify the following expression.
$$\frac { d } { d x } \int _ { 0 } ^ { x ^ { 2 } } \frac { d t } { t ^ { 2 } + 17 }$$
$$\frac { d } { d x } \int _ { 0 } ^ { x ^ { 2 } } \frac { d t } { t ^ { 2 } + 17 } =$$

Explanation:

Step1: Apply Leibniz Rule

The Leibniz rule for differentiation under the integral sign states that if \( F(x)=\int_{a(x)}^{b(x)} f(t) dt \), then \( F^\prime(x)=f(b(x))\cdot b^\prime(x)-f(a(x))\cdot a^\prime(x) \). Here, \( a(x) = 0 \), \( b(x)=x^{2} \), and \( f(t)=\frac{1}{t^{2}+17} \).
First, find \( a^\prime(x) \) and \( b^\prime(x) \). Since \( a(x) = 0 \), \( a^\prime(x)=0 \). For \( b(x)=x^{2} \), \( b^\prime(x) = 2x \).

Step2: Substitute into Leibniz Rule

Substitute into the formula: \( F^\prime(x)=f(x^{2})\cdot 2x - f(0)\cdot 0 \). Since \( f(t)=\frac{1}{t^{2}+17} \), \( f(x^{2})=\frac{1}{(x^{2})^{2}+17}=\frac{1}{x^{4}+17} \). And the second term is 0 because \( a^\prime(x) = 0 \). So \( F^\prime(x)=\frac{1}{x^{4}+17}\cdot 2x=\frac{2x}{x^{4}+17} \).

Answer:

\(\frac{2x}{x^{4}+17}\)