Question

If \(F(t)=\int_{2}^{3} \sin \left(x+t^{2}\right) d x\), find \(F^{\prime}(t)\).

Short answer

Answer: The derivative of the function \(F(t)\) with respect to \(t\) is \(F'(t) = \int_{2}^{3} \cos(x + t^2) \cdot 2t dx\).

Step-by-step solution

Find the partial derivative of \(f(x, t)\)

First, we need to find the partial derivative of \(f(x, t) = \sin(x + t^2)\) with respect to \(t\). Using the chain rule, we get: $$ \frac{\partial}{\partial t} f(x, t) = \frac{\partial}{\partial t} \sin(x + t^2) = \cos(x + t^2) \cdot \frac{\partial}{\partial t}(x + t^2) $$ Now, we find the derivative of \((x + t^2)\) with respect to \(t\): $$ \frac{\partial}{\partial t} (x + t^2) = 2t $$ Therefore, $$ \frac{\partial}{\partial t} f(x, t) = \cos(x + t^2) \cdot 2t $$

Integrate the partial derivative with respect to \(x\)

Now that we have the partial derivative, we can use the Leibniz rule to find \(F'(t)\): $$ F'(t) = \int_{2}^{3} \frac{\partial}{\partial t} f(x, t) dx = \int_{2}^{3} \cos(x + t^2) \cdot 2t dx $$

Write the final answer

We have found the derivative \(F'(t)\): $$ F'(t) = \int_{2}^{3} \cos(x + t^2) \cdot 2t dx $$ This is our final answer for the derivative of \(F(t)\).