Question
Find the derivative and simplify your answer. \[ \begin{array}{l}A(x)=\int_{2 x}^{1} \sec t d t \\ A^{\prime}(x)=[?][\quad] 2 x\end{array} \]
Ask by Warner Mathis. in the United States
Mar 22,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The derivative is \( A'(x) = -2\sec(2x) \).
Solution
Given
\[
A(x)=\int_{2x}^{1} \sec t \, dt,
\]
we differentiate with respect to \( x \) using the Leibniz rule for differentiating an integral with variable limits.
The Leibniz rule states that if
\[
F(x)=\int_{g(x)}^{h(x)} f(t) \, dt,
\]
then
\[
F'(x) = f(h(x))h'(x) - f(g(x))g'(x).
\]
Here, we identify:
- \( f(t)=\sec t \),
- Upper limit \( h(x)=1 \) so \( h'(x)=0 \),
- Lower limit \( g(x)=2x \) so \( g'(x)=2 \).
Substitute into the formula:
\[
A'(x)=\sec(1)\cdot 0-\sec(2x)\cdot 2 = -2\sec(2x).
\]
Thus, the derivative is
\[
A'(x)=-2\sec(2x).
\]
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Simplify this solution Extra Insights
To find the derivative \( A'(x) \) of the integral \( A(x) = \int_{2x}^{1} \sec t \, dt \), we can apply the Fundamental Theorem of Calculus along with the Chain Rule. First, we differentiate: \[ A'(x) = -\sec(2x) \cdot \frac{d}{dx}(2x) = -\sec(2x) \cdot 2 \] Thus, \[ A'(x) = -2 \sec(2x) \] So, the simplified answer is: \[ A'(x) = -2 \sec(2x) \] Therefore, we can fill in the blank as follows: \[ A^{\prime}(x) = -2 \sec(2x)[*] 2 x \]
