Example 2: Fundamental Theorem of Calculus Pt. 1

Compute the derivative:

$\frac{d}{\mathit{dx}}\underset{\mathit{sin}\left(x\right)}{\overset{7}{\int }}\frac{t}{1+t^2}\mathit{dt}$

Step 1: Ensure that the lower bound of your antiderivative is a constant, a , and the upper bound is an equation, x .

Here the bounds are not in the proper order. You will need to “pull a negative” to flip the order of the bounds.

$\underset{\mathit{sin}\left(x\right)}{\overset{7}{\int }}\frac{t}{1+t^2}\mathit{dt}$

$–\bullet \underset{7}{\overset{\mathit{sin}\left(x\right)}{\int }}\frac{t}{1+t^2}\mathit{dt}$

Step 2: Take the derivative of your antiderivative.

At this point you essentially cancel out the antiderivative and plug your upper bound equation , x , into the equation that was inside the antiderivative, $f\left(t\right)$.

Here you would replace all the t ’s in your equation with sin(x) ’s, the upper bound equation.

$\frac{d}{\mathit{dx}} –\bullet \underset{7}{\overset{\mathit{sin}\left(x\right)}{\int }}\frac{t}{1+t^2}\mathit{dt}\mathit{=}–\bullet \frac{\left(\mathit{sin}\right(x\left)\right)}{1+{\left(\mathit{sin}\right(x\left)\right)}^2}$

Step 3 (only if your equation is more than just an x ): If the upper bound is more than just a basic x , then you will need to multiply the equation you created in Step 2 , $f\left(x\right)$, by the derivative of the upper bound.

Here your upper bound is more than just an x , the upper bound is the equation sin(x) . You will need to multiply your result from Step 2 by the derivative of sin(x) , which is cos(x) .

$\frac{d}{\mathit{dx}} –\bullet \underset{7}{\overset{\mathit{sin}\left(x\right)}{\int }}\frac{t}{1+t^2}\mathit{dt}\mathit{=}–\bullet \frac{\left(\mathit{sin}\right(x\left)\right)}{1+{\left(\mathit{sin}\right(x\left)\right)}^2}\bullet \mathit{cos}\left(x\right)$

Final Result:

The derivative of the equation $F\left(x\right)=\underset{\mathit{sin}\left(x\right)}{\overset{7}{\int }}\frac{t}{1+t^2}\mathit{dt}$ would be $F\prime \left(x\right)=–\bullet \frac{\left(\mathit{sin}\right(x\left)\right)}{1+{\left(\mathit{sin}\right(x\left)\right)}^2}\bullet \mathit{cos}\left(x\right)$.