How to Calculate the Derivative of e^(x/2)^2?

[n3ll4f]

Hi

Can you please help me with this derivative:

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

 

[flatmaster]

You need to apply the chain rule three times.

 

[flatmaster]

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

e ^ ((x/2) ^2)

.Your Outside Function is..

e^(...)

The function inside that one is...

(...)^2

 

[n3ll4f]

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

 

[D H]

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

e ^ ((x/2) ^2)

That would be the standard interpretation. Without parentheses exponentiation evaluates right-to-left.[/QUOTE]

You need to apply the chain rule three times.

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?

 

[NoMoreExams]

Remember (and prove for yourself) that if

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

Then,

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

 

[n3ll4f]

Remember (and prove for yourself) that if

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

Then,

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

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

 

[NoMoreExams]

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

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 :)

 

[n3ll4f]

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

Thanks

 

[swraman]

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

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