Author Topic: Derivatives of the exponential function (Read 2403 times) Tweet Share

Damo17 · « **on:** November 18, 2008, 08:23:16 pm »

Find the derivative of e^-sqrt2x

I'm sure i'm missing something stupid. Any help would be much appreciated.

Glockmeister · « **Reply #1 on:** November 18, 2008, 08:32:03 pm »

$y = e^{-(2x)^{\frac{1}{2}}}$

For this question, you will need to use the chain rule, which you may or may not have heard about since it is not in the 1/2 Methods course. The chain rule states

$\frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx}$

$Let u = -(2x)^{\frac{1}{2}}$

$\frac{du}{dx} = -(2x)^{-\frac{1}{2}}$

$\therefore y = e^{u}$

$\frac{dy}{du} = e^{u}$

$\therefore \frac{dy}{dx} = -(2x)^{-\frac{1}{2}} . e^{u}$

$\frac{dy}{dx} = -\frac{e^{(2x)^{-\frac{1}{2}}}}{(2x)^{\frac{1}{2}}}$

$\frac{dy}{dx} = -\frac{e^{-\sqrt{2x}}}{\sqrt{2x}}$

Damo17 · « **Reply #2 on:** November 18, 2008, 08:35:30 pm »

It says the answer is -e^-sqrt2x/sqrt2x, how do I get that?

Quote from: Glockmeister on November 18, 2008, 08:32:03 pm

For this question, you will need to use the chain rule, which you may or may not have heard about since it is not in the 1/2 Methods course. The chain rule states

$\frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx}$

I'm using the quick method, does that matter?

Glockmeister · « **Reply #3 on:** November 18, 2008, 08:45:01 pm »

Quote from: Damo17 on November 18, 2008, 08:35:30 pm

It says the answer is -e^-sqrt2x/sqrt2x, how do I get that?

Quote from: Glockmeister on November 18, 2008, 08:32:03 pm
For this question, you will need to use the chain rule, which you may or may not have heard about since it is not in the 1/2 Methods course. The chain rule states

$\frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx}$

I'm using the quick method, does that matter?

If you can do it (and get the right answer) go for it. This is really the way for beginners to understand how you would go about doing this question.

Damo17 · « **Reply #4 on:** November 18, 2008, 08:48:48 pm »

Thanks Glockmeister so much, i spent ages on it.

Glockmeister · « **Reply #5 on:** November 18, 2008, 08:50:04 pm »

No problem mate.

the_head · « **Reply #6 on:** November 18, 2008, 09:08:13 pm »

Quote from: Glockmeister on November 18, 2008, 08:32:03 pm

$Let u = -(2x)^{\frac{1}{2}}$

$\frac{du}{dx} = -(2x)^{-\frac{1}{2}}$

This bit ain't right. du/dx needs to be divided by 2 as it is to the power of 0.5

Damo17 · « **Reply #7 on:** November 18, 2008, 09:12:46 pm »

^
Didn't see that. Care to share your solution?

the_head · « **Reply #8 on:** November 18, 2008, 09:20:57 pm »

Quote from: Damo17 on November 18, 2008, 09:12:46 pm

^
Didn't see that. Care to share your solution?

cbf doing the working (not much good at Latex lol) but did it by hand and this is what I got

$\frac{dy}{dx} = -\frac{e^{-\sqrt{2x}}}{2\sqrt{2x}}$

Damo17 · « **Reply #9 on:** November 18, 2008, 09:27:10 pm »

I guess i'll wait till tomorrow and ask my teacher. Thank you both anyway.

shinny · « **Reply #10 on:** November 18, 2008, 09:31:00 pm »

Quote from: the_head on November 18, 2008, 09:08:13 pm

Quote from: Glockmeister on November 18, 2008, 08:32:03 pm

$Let u = -(2x)^{\frac{1}{2}}$

$\frac{du}{dx} = -(2x)^{-\frac{1}{2}}$

This bit ain't right. du/dx needs to be divided by 2 as it is to the power of 0.5

But multiplying by the derivative of 2x cancels out the half. Glock's is fine.

Glockmeister · « **Reply #11 on:** November 18, 2008, 09:33:21 pm »

Quote from: the_head on November 18, 2008, 09:08:13 pm

Quote from: Glockmeister on November 18, 2008, 08:32:03 pm

$Let u = -(2x)^{\frac{1}{2}}$

$\frac{du}{dx} = -(2x)^{-\frac{1}{2}}$

This bit ain't right. du/dx needs to be divided by 2 as it is to the power of 0.5

huh?

$u = -(2x)^{\frac{1}{2}}$

$Let a = 2x$

$u = -a^{\frac{1}{2}}$

$\frac{du}{da} = \frac{-a^{\frac{-1}{2}}}{2}$

$\frac{da}{dx} = 2$

$\therefore \frac{du}{dx} = \frac{-a^{\frac{-1}{2}}}{2}.2$

$\frac{du}{dx} = \frac{-1}{(2x)^{\frac{1}{2}}}$

Damo17 · « **Reply #12 on:** November 18, 2008, 09:37:16 pm »

^
Yeah, now that i see it like that your right. Thanks. I've been doing maths all day, this shows maybe I should stop for the night.

Glockmeister · « **Reply #13 on:** November 18, 2008, 09:39:36 pm »

Quote from: Damo17 on November 18, 2008, 09:37:16 pm

^
Yeah, now that i see it like that your right. Thanks. I've been doing maths all day, this shows maybe I should stop for the night.

Yeah with these questions, it's often chain rule upon chain rule which is easy to make mistakes on.

the_head · « **Reply #14 on:** November 18, 2008, 09:40:58 pm »

Quote from: shinjitsuzx on November 18, 2008, 09:31:00 pm

Quote from: the_head on November 18, 2008, 09:08:13 pm
Quote from: Glockmeister on November 18, 2008, 08:32:03 pm

$Let u = -(2x)^{\frac{1}{2}}$

$\frac{du}{dx} = -(2x)^{-\frac{1}{2}}$

This bit ain't right. du/dx needs to be divided by 2 as it is to the power of 0.5

But multiplying by the derivative of 2x cancels out the half. Glock's is fine.

Can't believe I missed that. Stupid me. Those are the marks that are gonna cost me on this years exams

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Author Topic: Derivatives of the exponential function (Read 2403 times) Tweet Share

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