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I have the following practice problem that I'd like to know how to solve before taking my test, can someone explain what is necessary?

$$D_x \int_{0}^{2x} \left[15 \sqrt{2t^2 + 3t + 4} \right] dt$$

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Is that supposed to be $D_x$, the derivative with respect to $x$? – Brian M. Scott Dec 8 '12 at 22:30
Yes, I believe it should be. – DoesTheLimExist Dec 8 '12 at 22:34
Do you see what to do if the $2x$ were replaced by an $x$ (use the fundamental theorem of calculus)? Do that, but use the chain rule to account for the $2x$. – Potato Dec 8 '12 at 22:36

$$\frac{d}{dx}\,\int_{a(x)}^{b(x)}f(t)\,dt = f(b(x))\,b'(x) - f(a(x))\,a'(x)$$

In this case, $b(x)=2x$ and $a(x) = 0$.

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Forgive me, I'm not sure I understand the notation you've used. does $f(x, t)$ just mean plugging x and t into function f? – DoesTheLimExist Dec 8 '12 at 22:58
@DoesTheLimExist $f$ is a function of $x$ and $t$, yes. e.g. $f(x, t)=x^2+t^2$ then $f(1, 2)=1^2+2^2$ – Argon Dec 8 '12 at 23:17
I didn't downvote. I'm currently trying to understand all of what your equation shows. I'm not quite as advanced as your understanding of calculus :) – DoesTheLimExist Dec 8 '12 at 23:25
@DoesTheLimExist Does this make more sense? – Argon Dec 8 '12 at 23:38
Wouldn't that leave us with the answer: $15 \sqrt{2(2t)^2 + 3(2t) + 4} * 2 - 15 \sqrt{2(0)^2 + 3(0) + 4} * 0$? – DoesTheLimExist Dec 9 '12 at 0:00

You know from the fundamental theorem of calculus that $$\frac{d}{du}\int_0^uf(t)~dt=f(u)\;.\tag{1}$$ If $u=2x$ and $f(t)=15\sqrt{2t^2+3t+4}$, $(1)$ becomes


If $$F(x)=\int_0^{2x}15\sqrt{2t^2+3t+4}~dt\;,$$

you now know $\dfrac{dF}{du}$; how do you get $\dfrac{dF}{dx}$ from this?

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I see that you substituted u in, but how did you get $8x^2+6x+4$ out of that? – DoesTheLimExist Dec 8 '12 at 22:57
@DoesTheLimExist $u=2x$ – Américo Tavares Dec 8 '12 at 23:02
@DoesTheLimExist: $2u^2+3u+4=2(2t)^2+3(2t)+4=8t^2+6t+4$ – Brian M. Scott Dec 8 '12 at 23:02
@DoesTheLimExist: No: $(2t)^2=(2t)(2t)=4t^2$, and twice that is $8t^2$. – Brian M. Scott Dec 8 '12 at 23:22
Isn't $2(2t)^2 = 4t^2$ though? Aside from that, would $\frac{dF}{du} = \frac{dF}{dx}$, meaning would they be the same thing? – DoesTheLimExist Dec 8 '12 at 23:23

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