Find if .

## Solution

This is kind of a tricky problem. Obviously, if we need to find , we need to take the derivative. And since we are already given an explicit formula for* y* only in terms of *x*, it seems like we can just go ahead and take the derivative. But unfortunately, having an *x *both in the base and the exponent makes it a bit more complicated.

### So what can we do then?

Finding the derivative of this function is going to require a little trick that seems a little counter intuitive. ** What we need to do is actually take the natural log of both sides of this equation**. The reason for this is that it will help us get rid of the exponent and put this in a form we can work with more easily.

$$y=x^x$$

$$ln (y) =ln \big( x^x \big)$$

Now that we have done this we can use one of the basic *log* rules that you will want to remember.

#### 3 log rules to remember

Really quickly I want to list the three main *log *rules that you will want to remember. These come up frequently, so you will want to remember these.

$$log(ab) = log(a) + log(b)$$ $$log \bigg( \frac{a}{b} \bigg) = lob(a) – log(b)$$ $$log \Big( a^b \Big) = b \cdot log(a)$$

#### How do we apply this to our problem?

Let’s look back to our equation to see where we were.

$$ln (y) =ln \big( x^x \big)$$

Notice the right side of our equation looks a lot like the third *log *rule from above. Based on that third *log *rule, we can move the *x *in the exponent down in front of the *log *and multiply rather than having to deal with an exponent.

$$ln(y) = x \cdot ln(x)$$

The reason I think this seems a little counter intuitive is that we no longer have an explicit formula for *y*. **But now the right side of our equation will be easier to take its derivative.** So now let’s see what happens if we take the derivative of both sides of the equation with respect to *x*.

### Taking the derivative

The reason we need to take the derivative with respect to *x* is that the question asked us to find . The *dx* in the denominator is the indicator that tells us that we need to differentiate with respect to *x.* So let’s do that now.

$$\frac{d}{dx} ln(y) = \frac{d}{dx} \big[x \cdot ln(x) \big]$$

#### First the left side

Taking the derivative of the left side of the equation will require the use of the chain rule since *y* is a function of *x*. This was explained in a bit more detail in my implicit differentiation lesson. You will also use the fact that the derivative of *ln(x)* is .

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} \big[x \cdot ln(x) \big]$$

#### Then the right side

In order to take the derivative of the right side of this equation, we will need to use the product rule. As I did in the product rule lesson, we’ll first want to call one part of our function *f *and the other will be *g*.

$$f=x$$ $$g=ln(x)$$

Now we need to find* f’* and *g’* to use the product rule formula.

$$f’=1$$ $$g’=\frac{1}{x}$$

And lastly, we just need to plug these four pieces into the product rule formula.

$$\frac{1}{y} \cdot \frac{dy}{dx} = (x) \bigg( \frac{1}{x} \bigg) + (1) \big( ln(x) \big)$$

Now that we have taken the derivative of both sides of the equation, we just need to simplify the equation and solve for .

### Solving for *dy/dx*

$$\frac{1}{y} \cdot \frac{dy}{dx} = (x) \bigg( \frac{1}{x} \bigg) + (1) \big( ln(x) \big)$$

$$\frac{1}{y} \cdot \frac{dy}{dx} = 1 + ln(x)$$

And all we need to do from here is multiply both sides by *y *to isolate .

$$\frac{dy}{dx} = y \big( 1+ln(x) \big)$$

With most implicit differentiation problems this would be a perfectly fine place to stop and say we’ve reached our answer. Finding in terms of *x *and *y *is frequently the best we can do. But in this case, we can actually get our answer only in terms of *x *so that we have an explicit derivative of the original function.

The reason we’re able to do this is that our original function was an explicit formula for *y*. Since we know we can actually go to our formula for and replace the *y *with . So,

$$\frac{dy}{dx} = x^x \big( 1+ln(x) \big)$$

And now we can say we have reached our answer!

I just want to circle back to those 3 log rules I discussed above. **They can be very useful when taking derivatives of exponential functions**, or in some strange cases products and quotients. They can be used to rewrite complex functions in a way that would make their derivative easier to find, so always be sure to be aware of those and know how to use them to manipulate a function when needed.

As I always say, the best way to learn this stuff is practice, practice, practice! So check out some of my other lessons and problem solutions on derivatives. The more you work with these concepts, the better you’ll start to understand them.

If you can’t find the answer to your question or the topic you want to read about on my site, send me an email at jakesmathlessons@gmail.com and I’ll get back to you as soon as I can. **You can also use the form below to join my email list and I’ll send you my calculus 1 study guide!**