Derivative e^(x/2)^2

1. Dec 18, 2008

n3ll4f

Hi

e ^ (x/2) ^2

I will have an exame in 3 hours and i want to know how to resolve this derivative.

Thanks to all

2. Dec 18, 2008

flatmaster

You need to apply the chain rule three times.

3. Dec 18, 2008

flatmaster

It's ambiguious how you have it written. I assume you mean...

e ^ ((x/2) ^2)

e^(....)

The function inside that one is...

(.......)^2

4. Dec 18, 2008

n3ll4f

yes, that's right...
How it can be resolved?

5. Dec 18, 2008

D H

Staff Emeritus
That would be the standard interpretation. Without parentheses exponentiation evaluates right-to-left.[/QUOTE]
Only one application of the chain rule is needed (if the expression to be differentiated is $\exp((x/2)^2)$, that is). Do you really need the chain rule to compute the derivative of (x/2)^2=x^2/4?

6. Dec 18, 2008

NoMoreExams

Remember (and prove for yourself) that if

$$f(x) = e^{g(x)}$$

Then,

$$\frac{df}{dx} = e^{g(x)} \frac{dg}{dx}$$

7. Dec 18, 2008

n3ll4f

So the derivative of (x/2)^2 is x?

8. Dec 18, 2008

NoMoreExams

Pretty unlikely. Remember that someone already told you that if you did not want to use the chain rule to rewrite

$$\left(\frac{x}{2}\right)^{2} = \frac{x^2}{4}$$

If you do want to use the chain rule then remember that you would get

$$\left[\left(\frac{x}{2}\right)^{2}\right]^{'} = 2 \cdot \left(\frac{x}{2}\right) \cdot \frac{1}{2}$$

Since you have to differentiate "the inside" as well. That's what the chain rule is all about :)

9. Dec 18, 2008

n3ll4f

Ok, thanks.
You have help me a lot...
You're great..

Thanks

10. Dec 21, 2008

swraman

$$\frac{df}{dx} = e^{g(x)} \frac{dg}{dx}$$
That holds if it s e^(x). The more general form is slightly different:

f(x) = n^h(x) where n is any number
f'(x)= n^h(x) * ln(n) * h'(x)

when n = e, ln(n) = 1.

Even more generally, the derivatve of f(x) = r(x)^h(x) can be shown easily by the following proof:
f(x) = r(x)^h(x)
ln(f(x)) = ln(r(x)^h(x))
ln(f(x)) = h(x) ln(r(x))
f'(x)/f(x) = h(x)r'(x)/r(x) + h'(x)ln(r(x))
f'(x) = f(x) [h(x)r'(x)/r(x) + h'(x)ln(r(x))]

f'(x) = r(x)^h(x) [h(x)r'(x)/r(x) + h'(x)ln(r(x))]

Last edited: Dec 21, 2008