why is this derivative wrong?

**solidstatemath** · October 1st 2010, 11:08 AM

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

**SpringFan25** · October 1st 2010, 11:13 AM

fine so far, but you haven't finished.

$\displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right)$

$\displaystyle \frac{d}{dx} \left( \left[ 0.25 e^{-4t} \right]^{x^3}_0 \right)$

$\displaystyle \frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25e^0 \right)$

$\frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25 \right)$

now differentiate with respect to x.

$0.25 \frac{d}{dx} \left( e^{-4x^3} \right) - 0$

...

...

**SpringFan25** · October 1st 2010, 11:18 AM

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 )

**solidstatemath** · October 1st 2010, 11:46 AM

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?

**SpringFan25** · October 1st 2010, 11:50 AM

Yes on the factoring, but the derivative of 1 is 0.

$\frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25 \right)$

$0.25 \frac{d}{dx} \left( e^{-4x^3} - 1 \right)$

$0.25 \left( \frac{d}{dx} \left( e^{-4x^3} \right) - \frac{d}{dx}\left(1 \right) \right)$

$0.25 \left( \frac{d}{dx} \left( e^{-4x^3} \right) - 0 \right)$

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 ^^

**solidstatemath** · October 1st 2010, 11:59 AM

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"?

**SpringFan25** · October 1st 2010, 12:01 PM

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 $\frac{d}{dx}\left( f(x) \right)$ means. It does not mean that there is some fraction $\frac{d}{dx}$ being multuiplied by a function, f(x).

**solidstatemath** · October 1st 2010, 12:06 PM

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 ).

**SpringFan25** · October 1st 2010, 12:13 PM

and what does the textbook say the answer is?

**solidstatemath** · October 1st 2010, 12:19 PM

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.

**Plato** · October 1st 2010, 12:22 PM

I think that you are missing the point of the question.

It is well known that if $f$ is continuous on $[a,b]$ and $\displaystyle g(x) = \int\limits_a^x {f(t)dt}$ then $g'(x)=f(x).$

So with the aid of the chain rule $\displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right)=(3x^2)e^{-4x^3}$

**SpringFan25** · October 1st 2010, 12:31 PM

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 $\displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right)$

which i will write as $\frac{d}{dx}something$

From the previous discussion, you should agree that
"something" = $0.25e^{-4x^3} - 0.25$

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 $\displaystyle \left(0.25e^{-4x^3} - 0.25\right)$

$=0.25 \times \frac{d}{dx}(-4x^3} \times e^{-4x^3} -0$ chain rule

$=0.25 \times -12x^2 \times e^{-4x^3}$

$= -3x^2 \times e^{-4x^3}$

**solidstatemath** · October 3rd 2010, 12:34 PM

Okay just did this again from scratch, this is what I got:

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

What do you guys think?

**Archie Meade** · October 3rd 2010, 01:23 PM

Originally Posted by solidstatemath

Okay just did this again from scratch, this is what I got:

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

What do you guys think?

Yes, that's the integration complete.
Differentiate that and have another look at Plato's post.

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