Quick Question about taking the Derivative of An Expression from Fellow User

Question

Since Stackexchange doesn't allow comments, I though I'd ask a quick question about an integral here from the post Trapezoid rule error analysis .

$g(t)=\int_{p-t}^{p+t} f(x) \,dx−t({f(p−t)+f(p+t))}$

$g′(t)=−t({f′(p+t)−f′(p−t)})$

This is the function and it's derivative and I can't seem to understand where the f(x) term went and the second part of the expression.

For the first term (which is an integral) the derivative is $f(p+t) +f(p-t) $ (via Fundamental Theorem of Calculus) and the next terms are differentiated using product rule. Some cancellation occurs and the final answer is as given. — Paramanand Singh, Jul 11 '17 at 12:30

score 1 · Accepted Answer · answered Jul 11 '17 at 12:36

1

Let $F(x)$ be the anti-derivative of $f(x)$. Then $g(t)=F(p+t)-F(p-t)-tf(p-t)-tf(p+t)$.

Therefore $g^{'}(t)=f(p+t) - f(p-t)\times(-1) -tf^{'}(p-t)\times(-1) -f(p-t) -tf^{'}(p+t) - f(p+t)$

Cancel out the terms and you have the answer

answered Jul 11 '17 at 12:36

praveen kr

I have multiplied the term with $(-1)$ in the end just to show the chain rule explicitly. Sorry if it is not readable. – praveen kr Jul 11 '17 at 14:39
It's my mistake then. Perhaps I was in haste. Will delete my previous comment. – Paramanand Singh Jul 11 '17 at 15:10

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