# Thread: Derivative of a function

1. ## Derivative of a function

How do you take the derivative of f(x) = arctan(2^x)?

2. well, all you have to know is this:

$\displaystyle \frac{dy}{dx}(arctan(x)) = \frac{1}{1 + x^2}$

Then, you have to apply the chain rule to find the derivative of the inside function $\displaystyle x^2$.

So, here is your entire problem.

$\displaystyle f(x) = arctan(x^2)$
$\displaystyle f'(x) = \frac{1}{1+(x^2)^2}*2x$
$\displaystyle f'(x) = \frac{2x}{1+x^4}$

3. Sorry, I put down the wrong problem.

The correction question is:

f(x) = arctan(2^x)

4. Same principle. Let $\displaystyle u=2^x$.

Thus $\displaystyle \frac{d}{dx} \arctan(u) = \frac{1}{1+u^2} \times \frac{du}{dx}$

Do you know how to find the derivative of u?

5. So do you take the natural log of both sides now?

So do you take the natural log of both sides now?
No. That is not necessary.
For any $\displaystyle a>0$ the derivative of $\displaystyle a^x$ is $\displaystyle a^x \ln (a)$.

7. Yes, but to derive that rule you use natural logs.

$\displaystyle u=2^x$

$\displaystyle \ln(u)=x\ln(2)$

$\displaystyle \frac{1}{u} \times \frac{du}{dx}=\ln(2)$

$\displaystyle \frac{du}{dx}=u\ln(2)$