Question

Find dy/dx of the integral from x^3 to 5 of the function (cos t)/ (t^2 + 2) dt

Answer 1

Find \(\large\color{black}{\displaystyle\frac{dy}{dx}}\)

if: \(\large\color{black}{\displaystyle f(x)=\int\limits_{5}^{x^3}\frac{\cot (t)}{t^2+2}~ dt}\)


well, going by a fundamental theorem of calculus, it would be an integral from 5 to x^3,
So, you plug in x^3 and 5 (the upper and lower limits).

then, as you differentiate f(5) becomes 0, and f(x^3) becomes what is inside the integral, but with a chain rule for x^3.

Answer 2

where S'=s

\(\large\color{black}{\displaystyle f(x)=\int\limits_{a(x)}^{b(x)} s(t)~dt}\)


\(\large\color{black}{\displaystyle f(x)=\int\limits_{a(x)}^{b(x)} s(t)~dt=S(t)~{{\huge |}_{ a(x)}^{b(x)}}}\)


\(\large\color{black}{\displaystyle f(x)=\int\limits_{a(x)}^{b(x)} s(t)~dt=S(b(x))-S(a(x))}\)

Answer 3

now, in your case a(x) is zero, so when you differentiate S(a(x)) is it out.

Answer 4

so when a(x) is constant.

\(\large\color{black}{\displaystyle f(x)=\int\limits_{a(x)}^{b(x)} s(t)~dt=S(b(x))-S(a(x))}\)

\(\large\color{black}{\displaystyle f'(x)=s(b(x))~\times b'(x)}\)

Answer 5

that is the chain rule.

Answer 6

when I said
`now, in your case a(x) is zero, so when you differentiate S(a(x)) is it out.`
I meant:
`now, in your case a(x) is constant so when you differentiate S(a(x)) is it out.`

Answer 7

So can you explain how the chain rule works in a case like this? I know the derivative is 3x^2 but how do I use that here?

Answer 8

oh it's cosine not cotangent, but technic is same

Answer 9

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos t}{t^2+2}~dt}\)
here you know that derivative with respect to t, is just:

\(\large\color{slate}{\displaystyle\frac{\cos t}{t^2+2}}\) correct?

Answer 10

Right

Answer 11

And so, when you integrate, plug in x^3 -- the upper limit.

(don't worry about the lower limit 5, because that is going to be a constant and is going to be zero as you differentiate)

and you get that the derivative 9with respect to x, after you plugged in x^3 for t, as the upper limit) is same,

\(\large\color{slate}{\displaystyle\frac{\cos (x^3)}{(x^3)^2+2}}\) . however...

Answer 12

\(\large\color{slate}{\displaystyle\frac{\cos (x^3)}{(x^3)^2+2}}\) is just a derivative without a chain rule, but to differentiate that integral when you have x^3, you would need the chain rule, and so, what we do is:

Answer 13

\(\large\color{slate}{\displaystyle\frac{\cos (x^3)}{(x^3)^2+2} \times \left( \frac{d}{dx}~x^3\right)}\)

Answer 14

Oh okay so its the whole thing times the derivative. I was trying to multiply that in both numerator and denominator and it canceled out and ugh. Alright that explanation really helped. Thanks so much!

Answer 15

yes. (for a lower limit would be same, but negative)

Answer 16

ywe