# Applying the fundamental theorem of calculus to find a derivative.

From Logan, D. (2017). A first course in differential equations. Springer.

What are the steps to simplify that derivative?

EDIT: As pointed out in the comments there is likely a typo in the answer to 11. With that in mind it is actually quite simple to solve this. As I have done below. The initial derivative of $$y(t)$$ should read....

$$y'(t) = e^{-t^2}e^{t^2}-2te^{-t^2} \int_0^t e^{s^2}ds$$

• Could there be a typo in 11, i.e., an $e^{-t^2}$ accidentally became $e-t^2$? No one can tell except you by actually using the product rule as suggested … Commented Jan 4, 2022 at 8:05
• If you wrote the equations in MathJax, the question would look much better :) Commented Jan 4, 2022 at 8:20
• @Hermis14 Then again, in that case we would have suspected that The typo was introduced by Angus and not the original Commented Jan 4, 2022 at 9:06
• Yes thank you now it is quite simple to resolve the steps taken. I was very confused with the initial statement and assumed I had done something horribly wrong. Commented Jan 4, 2022 at 17:47
• Please, use descriptive titles. "Understanding the following steps..." says nothing about the subject of the question. Commented Jan 4, 2022 at 19:28

$$y'(t) = e^{-t^2}e^{t^2}-2te^{-t^2} \int_0^t e^{s^2}ds$$ $$y'(t) = \frac{e^{t^2}}{e^{t^2}}-2t\underbrace{e^{-t^2} \int_0^t e^{s^2}ds}_{=y}$$ $$y'(t) = 1- 2ty$$