Question

What is `dy/dx` when `y=(arctanx)^(tanx) , x>0?`

Answer 1

`dy/dx = y[sec^2(x)ln[arctan(x)] + (tan(x)/(arctan(x)(1+x^2)))]`

Alternatively:
`dy/dx = [arctan(x)^tan(x)][sec^2(x)ln[arctan(x)] + (tan(x)/(arctan(x)(1+x^2)))]`

is equally valid.

Explanation:

For this problem, notice that you have a variable in both your base and exponent. For this reason, you cannot use either the exponent rule (like you would for `a^x`) or the power rule, as much as you may be tempted to.

Hence, you'll need to do a nifty bit of algebra to solve this. The first step is to take the natural log of both sides, as follows:

`ln(y) = ln([arctan(x)]^tan(x))`

Now, you can use a property of logarithms (I have a video on this, if you need a refresher) to bring down the exponent to the front, as follows:

`ln(y) = tan(x)ln([arctan(x)])`

Now, this is simply a chain rule - product rule problem. The only difference is that you have a `ln(y)` on the left side of the equation, meaning that you'll also need to invoke implicit differentiation there.

All said, after you've taken your derivative (I'm going to assume you know how to do the product rule for the right hand side), you should end up with:

`1/y(dy/dx) = sec^2(x)ln[arctan(x)] + (tan(x)/(arctan(x)(1+x^2)))`

Note that there's a `1/y` on the left hand side of the equation. This is merely a result of implicit differentiation.

There's just one final step: solve for `dy/dx`. It's pretty straightforward; all you need to do is multiply both sides of the equation by `y`. You will now be left with:

`dy/dx = y[sec^2(x)ln[arctan(x)] + (tan(x)/(arctan(x)(1+x^2)))]`

This is a perfectly acceptable final answer. Now, if you'd like to take this one step further and have your final answer in terms of `x`'s, then all you need to do is plug in the value for `y` (i.e. your original function), which should leave you with:

`dy/dx = [arctan(x)^tan(x)][sec^2(x)ln[arctan(x)] + (tan(x)/(arctan(x)(1+x^2)))]`

For this particular problem, plugging in for `y` actually turns out to work out pretty nicely. However, in more complicated problems (especially when you have to deal with quotients), things can get pretty messy, so the general rule I have is to leave the final answer in terms of both `y` and `x` unless instructed otherwise (or if plugging in `y` works out nicely like it does here).

That's all! If you'd like some further assistance with this concept, I have a couple of videos you can check out:

[Logarithmic Differentiation Theory Video ()
[Logarithmic Differentiation Practice Problem Video ()

Hope that helps :)