See attachment. It should be right but I'm wrong.

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- Oct 1st 2010, 11:08 AMsolidstatemathwhy is this derivative wrong?
See attachment. It should be right but I'm wrong.

- Oct 1st 2010, 11:13 AMSpringFan25
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 $

...

... - Oct 1st 2010, 11:18 AMSpringFan25
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 ) - Oct 1st 2010, 11:46 AMsolidstatemath
Ok I understand what you did in the first four lines. I don't get the part where you say differentiate with respect to x and it goes "- 0 " at the end. Shouldn't it be -1 if the 0.25 was factored out like that?

- Oct 1st 2010, 11:50 AMSpringFan25
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 ^^ - Oct 1st 2010, 11:59 AMsolidstatemath
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"? - Oct 1st 2010, 12:01 PMSpringFan25
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). - Oct 1st 2010, 12:06 PMsolidstatemath
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 ). - Oct 1st 2010, 12:13 PMSpringFan25
and what does the textbook say the answer is?

- Oct 1st 2010, 12:19 PMsolidstatemath
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. - Oct 1st 2010, 12:22 PMPlato
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} $ - Oct 1st 2010, 12:31 PMSpringFan25
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} $ - Oct 3rd 2010, 12:34 PMsolidstatemath
Okay just did this again from scratch, this is what I got:

1/4 [(-e^-4x^3) + 1]

What do you guys think? - Oct 3rd 2010, 01:23 PMArchie Meade