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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: kenhung123 on June 11, 2010, 12:52:04 pm

Title: Differentiating e^x
Post by: kenhung123 on June 11, 2010, 12:52:04 pm: So when we differentiate functions in the form $f(x)=e^{x}$ does it mean $f'(x)=e^{x}$ because of the chain rule?
$So y=e^{u}$ where $u=x^{2}+4$
Does it mean the derivative of exponentials are always itself but because of the chain rule, we may have a term to multiply?
Title: Re: Differentiating e^x
Post by: Aqualim on June 11, 2010, 01:06:57 pm: In General;

$f(x)= e^x$
$let x=u$ in this case
$\therefore f'(x)=f'(u)e^u$

Now this works with the original aswell cause we know the derivative of $x$ becomes 1 so basically you are just multiplying $e^x$ by 1 which leaves it as it is.

So for the question you have above you first find the derivative of $u=x^2+4$ then you just place it out the front of $e^u$ . Which becomes;

$2xe^{x^2+4}$
Title: Re: Differentiating e^x
Post by: tcg93 on June 11, 2010, 01:13:04 pm: Quote from: kenhung123 on June 11, 2010, 12:52:04 pm
So when we differentiate functions in the form $f(x)=e^{x}$ does it mean $f'(x)=e^{x}$ because of the chain rule?
$So y=e^{u}$ where $u=x^{2}+4$
Does it mean the derivative of exponentials are always itself but because of the chain rule, we may have a term to multiply?

Yep. Pretty much. I just remember it as
$f'(x)=$ (Derivative of the power) x $f(x)$
where in this case the Derivative of the Power is 1.
Title: Re: Differentiating e^x
Post by: tcg93 on June 11, 2010, 01:14:07 pm: Quote from: tcg93 on June 11, 2010, 01:13:04 pm
Quote from: kenhung123 on June 11, 2010, 12:52:04 pm
So when we differentiate functions in the form $f(x)=e^{x}$ does it mean $f'(x)=e^{x}$ because of the chain rule?
$So y=e^{u}$ where $u=x^{2}+4$
Does it mean the derivative of exponentials are always itself but because of the chain rule, we may have a term to multiply?

Yep. Pretty much. I just remember it as
$f'(x)=$ (Derivative of the power) x $f(x)$
where in this case the Derivative of the Power is 1.

Note that is just for exponentials, not polynomials
Title: Re: Differentiating e^x
Post by: TyErd on June 11, 2010, 01:46:19 pm: Quote from: tcg93 on June 11, 2010, 01:13:04 pm
Quote from: kenhung123 on June 11, 2010, 12:52:04 pm
So when we differentiate functions in the form $f(x)=e^{x}$ does it mean $f'(x)=e^{x}$ because of the chain rule?
$So y=e^{u}$ where $u=x^{2}+4$
Does it mean the derivative of exponentials are always itself but because of the chain rule, we may have a term to multiply?

Yep. Pretty much. I just remember it as
$f'(x)=$ (Derivative of the power) x $f(x)$
where in this case the Derivative of the Power is 1.

I also do it that way but Im just wondering would I lose marks in the exams because usually my answers are one liners with no working out.
Title: Re: Differentiating e^x
Post by: Ilovemathsmeth on June 11, 2010, 02:05:16 pm: I absolutely love differentiation.

Okay, that said.

Another common trap people fall into is differentiating exponentials of other bases. Keep in mind that the above formula of (derivative of u) $f^u$ only works for base e. If you've got base 10, or 2 or whatever, you HAVE to use the change of base rule to make it base e, and THEN you can apply the above rule. This came up on a SAC once and a LOT of people got it wrong :)
Title: Re: Differentiating e^x
Post by: the.watchman on June 11, 2010, 02:21:31 pm: Quote from: Ilovemathsmeth on June 11, 2010, 02:05:16 pm
Another common trap people fall into is differentiating exponentials of other bases. Keep in mind that the above formula of (derivative of u) $f^u$ only works for base e. If you've got base 10, or 2 or whatever, you HAVE to use the change of base rule to make it base e, and THEN you can apply the above rule. This came up on a SAC once and a LOT of people got it wrong :)

Oh, and might I add that the same applies to logs, where $\frac{d}{dx}(log_e (x)) = \frac{1}{x}$
This only works for base e, and so for other bases, use the change base formula :)
Title: Re: Differentiating e^x
Post by: kenhung123 on June 11, 2010, 04:02:30 pm: Could you give an example?
Title: Re: Differentiating e^x
Post by: tcg93 on June 11, 2010, 04:08:00 pm: Quote from: kenhung123 on June 11, 2010, 04:02:30 pm
Could you give an example?

$(log4(x))$

= $(loge(x))/(loge(4))$

Now $(1)/(loge(4))$ is a constant

Therefore the derivative is $(1)/(xloge(4))$
Title: Re: Differentiating e^x
Post by: the.watchman on June 11, 2010, 04:09:35 pm: Ok:

$\frac{d}{dx}(log_2 (x))$

$=\frac{d}{dx}(\frac{log_e (x)}{log_e (2)})$

$=\frac{d}{dx}(\frac{1}{log_e (2)} \cdot log_e (x))$ (loge(2) is a constant)

$=\frac{1}{log_e (2)} \cdot \frac{1}{x}$

$=\frac{1}{x \cdot log_e (2)}$

Try an exponential one yourself :)
Title: Re: Differentiating e^x
Post by: kenhung123 on June 11, 2010, 07:01:33 pm: Hmm??? Sorry won't constants dissapear?
Title: Re: Differentiating e^x
Post by: the.watchman on June 11, 2010, 07:48:09 pm: Only when they're by themselves: like x^2+1 => 2x

But in this scenario(as a coefficient), it's like kx^2 => 2kx (if you know what I mean)
Title: Re: Differentiating e^x
Post by: stonecold on June 11, 2010, 07:54:13 pm: This thread is going to help with my SAC coming up on Thursday. Thanks. :D
Title: Re: Differentiating e^x
Post by: Ilovemathsmeth on June 12, 2010, 04:35:20 pm: Exactly. Please use change of base rule for logarithms to differentiate; this came up on a SAC and there were a LOT of people who incorrectly differentiated and lost marks as a result.