I have attached a question that I came across in trying to understand the fundamental theorem of calculus. The solution (highlighted with an arrow). I have difficulty understanding why the chain rule was not applied in the final step (highlighted in the rectangle). My initial "incorrect" solution was -6(3x+1) can someone please clarify why the chain rule is not applicable to this equation?

enter image description here

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    $\begingroup$ Because there is no composition, the variable is just $x$ and it's in the integral limit, it's not a function of $x$. $\endgroup$ – Adam Hughes Jul 24 '14 at 21:02
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    $\begingroup$ The Chain Rule would have to be applied to find the derivative of, for example, $\int_1^{x^3} (3t+1)^2\,dt$. $\endgroup$ – André Nicolas Jul 24 '14 at 21:03
  • $\begingroup$ Thank you for the comments. André, why does changing the upper bound of the integral change how the derivative is solved? sorry for keeping on asking I am trying to teach myself calculus $\endgroup$ – MacUser Jul 24 '14 at 21:08
  • $\begingroup$ The chain rule is already applied. Please clarify why you think it isn't applied. $\endgroup$ – Gina Jul 24 '14 at 21:30

The chain rule is: $$\begin{align} \frac{\operatorname{d}}{\operatorname{d} x}\int_c^{g(x)} f(t) \operatorname{d} t & = \frac{\operatorname{d}g(x)}{\operatorname{d} x}\cdot \frac{\operatorname{d}}{\operatorname{d} g(x)} \int_c^{g(x)} f(t) \operatorname{d} t \\ & = g'(x)\cdot f(g(x))\end{align}$$

In your example $g(x)=x$ so $g'(x)=1$, and thus by substitution, leaves just the fundamental theorem:

$$\frac{\operatorname{d}}{\operatorname{d} x}\int_c^{x} f(t) \operatorname{d} t = f(x)$$


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