1. ## Derivative of 2^(xtanx)?

I don't even know where to begin on this one. It's not (xtanx)2^(xtanx-1) is it?

2. Hello, satx!

You're expected to know this differentiation formula:

. . $\text{If }\,f(x) \;=\;b^{u},\,\text{ then: }\:f'(x) \;=\;b^u\,u' \,\ln b$

$f(x) \;=\;2^{x\tan x}$

We have: . $f'(x) \;=\;2^{x\tan x}\cdot(x\sec^2x + \tan x)\cdot \ln 2$

3. Originally Posted by satx
I don't even know where to begin on this one. It's not (xtanx)2^(xtanx-1) is it?
No, it isn't. The "power law" only applies when the base is a variable and the power a constant.

You need to know, first, that the derivative of $a^x$ is $ln(a) a^x$. Further, by the chain rule, the derivative of $a^u(x)$ is $ln(a) a^{u(x)}\frac{du}{dx}$.

In this case, of course, u(x)= x tan(x). Use the product rule to find du/dx.

(Once again, Soroban beats me by 1 minute!!!)

4. Originally Posted by satx
I don't even know where to begin on this one. It's not (xtanx)2^(xtanx-1) is it?
Use:
The Exponential Rule

$\boxed{\frac{d}{dx}b^x = b^x \cdot \ln{b}}$

And the chain rule in the exponential rule:

$\frac{d}{dx}b^{f(x)} = b^{f(x)} \cdot \ln{b} \cdot f'(x)$

In your problem, $b = 2$ and $f(x) = x \tan(x)$

Now use the product rule to find $f'(x)$:

$f'(x) = x\sec^2(x) + \tan(x) \cdot 1$

So the answer is $2^{x\tan(x)}\cdot \ln2 \cdot (x\sec^2(x) + \tan(x))$

Hope that helps

Mathemagister

PS Sorry for the post taking so long and overlapping others' posts (my internet disconnected)

5. Merci beaucoup a tous!

6. Some of the posters in this thread seem to think that calculus is an exercise in the memorisation of a fist full of "rules".

Here you need the definition of logarithms and a bit of manipulation:

$a^b=e^{\ln(a)b}$

Then the rest follows from the systematic application of the product and chain rules and a table of basic derivatives.

Brain storage space is to precious to waste on large list of rules, understanding acts as a compression algorithm, the more you understand the less you need to remember.

CB

7. Originally Posted by CaptainBlack
Some of the posters in this thread seem to think that calculus is an exercise in the memorisation of a fist full of "rules".

Here you need the definition of logarithms and a bit of manipulation:

$a^b=e^{\ln(a)b}$

Then the rest follows from the systematic application of the product and chain rules and a table of basic derivatives.

Brain storage space is to precious to waste on large list of rules, understanding acts as a compression algorithm, the more you understand the less you need to remember.

CB
Yes, but it is much easier to teach that way. Even though one understands the underlying meaning, it is more comprehensible for the person in doubt if they are taught in terms of the rules, which are a sort of the language in which we explain how to do a problem which is then translated back by the other person (hopefully) in such a way that they understand the underlying meaning.

8. Originally Posted by mathemagister
Yes, but it is much easier to teach that way. Even though one understands the underlying meaning, it is more comprehensible for the person in doubt if they are taught in terms of the rules, which are a sort of the language in which we explain how to do a problem which is then translated back by the other person (hopefully) in such a way that they understand the underlying meaning.
I'm sorry but I don't agree, for instance I know there is a quotient rule, but do I know what it is? No because it is actually easier to use the product rule once the chain rule is internalised. Rule that are only occasionally used are more trouble that they are worth.

Also rule based rote learning gives the wrong idea to the student of what mathematics is. I don't care if the student manages to clear artificial hurdles what I do care about is that they learn to appreciate what maths is about (beauty and pattern) and can use what they do know in novel situations.

CB