0
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.