MATH101:Calculus I-2024 First All Sections General
If , then

Ask by Morgan Johnston. in Jordan

Jan 06,2025

Question

MATH101:Calculus I-2024 First All Sections General If $$ F(x)=\int_{2 x}^{x^{2}} \sqrt{t^{2}+1} d t $$, then $$ F^{\prime}(x)= $$

UpStudy AI · Accepted Answer

To find the derivative $$ F^{\prime}(x) $$ of the function $$ F(x) = \int_{2x}^{x^{2}} \sqrt{t^{2}+1} \, dt $$, we can apply the **Leibniz rule** for differentiation under the integral sign. The Leibniz rule states:

$$
F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)
$$

where:
- $$ f(t) = \sqrt{t^{2} + 1} $$
- $$ a(x) = 2x $$ with derivative $$ a'(x) = 2 $$
- $$ b(x) = x^{2} $$ with derivative $$ b'(x) = 2x $$

Applying the rule to our function:

$$
F'(x) = \sqrt{(x^{2})^{2} + 1} \cdot (2x) - \sqrt{(2x)^{2} + 1} \cdot 2
$$

Simplifying the expression inside the square roots:

$$
F'(x) = 2x \sqrt{x^{4} + 1} - 2 \sqrt{4x^{2} + 1}
$$

**Final Answer:**

After simplifying, the derivative is twice x√(x⁴ + 1) minus twice √(4 x² + 1). Thus,

F ′(x) = 2x √(x⁴+1) − 2 √(4x²+1)
$$ F^{\prime}(x) = 2x \sqrt{x^{4} + 1} - 2 \sqrt{4x^{2} + 1} $$

MATH101:Calculus I-2024 First All Sections General
If , then

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MATH101:Calculus I-2024 First All Sections General If , then