See attachment. It should be right but I'm wrong.
fine so far, but you haven't finished.
$\displaystyle \displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right) $
$\displaystyle \displaystyle \frac{d}{dx} \left( \left[ 0.25 e^{-4t} \right]^{x^3}_0 \right) $
$\displaystyle \displaystyle \frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25e^0 \right) $
$\displaystyle \frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25 \right) $
now differentiate with respect to x.
$\displaystyle 0.25 \frac{d}{dx} \left( e^{-4x^3} \right) - 0 $
...
...
also, according to wikipedia, the answer can be written down directly:
Differentiation under the integral sign - Wikipedia, the free encyclopedia
But you should use whatever method your teacher is expecting.
Edit: Ignore this, not applicable here. (would be if the upper limit was x instead of x^3 )
Yes on the factoring, but the derivative of 1 is 0.
$\displaystyle \frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25 \right) $
$\displaystyle 0.25 \frac{d}{dx} \left( e^{-4x^3} - 1 \right) $
$\displaystyle 0.25 \left( \frac{d}{dx} \left( e^{-4x^3} \right) - \frac{d}{dx}\left(1 \right) \right) $
$\displaystyle 0.25 \left( \frac{d}{dx} \left( e^{-4x^3} \right) - 0 \right) $
Note There MAY be a mistake in my working somewhere else, as i dont think the answer that this will produce is consistent with what wikipedia says. However wikipedia math articles routinely have mistakes in them, so we will see ^^
ok so the - 0 at the end isn't needed leaving us with just 0.25 d/dx * e^-4x^3 and isn't that what I wrote or should I ad "dt"?
This would give (0.25 d/dx * e^-4x^3) * dt
or is it because they ask for the derivative of d/dx that I must replace "t" by "x^3"?
I dont see that in your screenshot. Even if it was what you meant, you haven't answered the question. you have to actually DO the differenciation.
the answer is not d/dx "times" something. The answer is the derivative of 0.25 e^(-4x^3)
NB
Make sure you understand what the notation $\displaystyle \frac{d}{dx}\left( f(x) \right) $ means. It does not mean that there is some fraction $\displaystyle \frac{d}{dx}$ being multuiplied by a function, f(x).
I don't understand.
The way I read your line 4 is like this: 0.25 ( d/dx ( e^-4x^3 ) -0 )
When I see this I read 0.25 and then the derivative of what's inside. If that is it then it should be e^-4x^3 * -12x^2 and we would get 0.25 * ( e^-4x^3 * -12x^2 ).
There is none. The tutor I went to see gave me this to work with:
integral sign = S
S (e^ax) * dx = 1/a * e^ax
so I got: S (e^-4t)dt = -1/4 * e^-4t
Then I think I sub t for x^3 but I'm not sure. I really don't know what to do after the sub. Do I apply F(b) - F(a) because I'm really lost. I'm so used to doing the F(b)-F(a) thing that when they ask this I just end up totally lost.
I think that you are missing the point of the question.
It is well known that if $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and $\displaystyle \displaystyle g(x) = \int\limits_a^x {f(t)dt} $ then $\displaystyle g'(x)=f(x).$
So with the aid of the chain rule $\displaystyle \displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right)=(3x^2)e^{-4x^3} $
0.25 * ( e^-4x^3 * -12x^2 ) was correct.
You already did F(b) - F(a) right at the start to obtain this answer.
Starting over:
You want to find $\displaystyle \displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right) $
which i will write as $\displaystyle \frac{d}{dx}something$
From the previous discussion, you should agree that
"something" = $\displaystyle 0.25e^{-4x^3} - 0.25$
you get this by doing the integral in the normal way, including the F(b) - F(a) step at the end.
NOW start thinking about the derivative of "something", using the chain rule.
differenciate $\displaystyle \displaystyle \left(0.25e^{-4x^3} - 0.25\right) $
$\displaystyle =0.25 \times \frac{d}{dx}(-4x^3} \times e^{-4x^3} -0$ chain rule
$\displaystyle =0.25 \times -12x^2 \times e^{-4x^3} $
$\displaystyle = -3x^2 \times e^{-4x^3} $