Question

In Exercises \(1-20,\) find \(d y / d x\). $$y=\int_{5 x^{2}}^{25} \frac{t^{2}-2 t+9}{t^{3}+6} d t$$

Short answer

\(d y/d x = -[25x^{4}-10x^{2}+9)(10 * x)/((125x^{6}+6)]\)

Step-by-step solution

Applying Negative Fundamental Theorem of Calculus

Because the lower limit is the function of x, we need to apply the negative version of the Fundamental Theorem of Calculus. Therefore, \(d y/d x=-\frac{d}{d x}\left[\int_{5 x^{2}}^{25} \frac{t^{2}-2 t+9}{t^{3}+6} d t\right]\)

Apply Chain Rule

The chain rule needs to be applied next to differentiate not just the integral itself, but also the function in the limit of integration: \(d y/d x=-\frac{d \int_{5 x^{2}}^{25} \frac{t^{2}-2 t+9}{t^{3}+6} d t}{d(5 x^{2})} * \frac{d (5 x^{2})}{d x} \)

Application of FFTC and Simplification

Now we can apply the Fundamental Theorem of Calculus, which states that the derivative of an integral of a function is simply that function evaluated at the boundaries of the integral. Hence, simplification of the expression can be done as follows: \(d y/d x = -[\frac{(5x^{2})^{2}-2(5x^{2})+9}{(5x^{2})^{3}+6}*2*5x]\), this further simplifies to \(d y/d x = -[25x^{4}-10x^{2}+9)(10 * x)/((125x^{6}+6)]\).