# The Derivative, Part 8

Now let’s do some more examples using not only the chain rule, but using a combination of the rules we have covered.

Let me start with an example that illustrates the “chainyness” of the chain rule. Let $f( x) =\text{sin}\left(\sqrt{x^{2} +2x-7}\right)$

Notice that there are three operations at work here: the sine, the square root and the polynomial. Referring back to the previous post, which is the outermost function? It’s the sine as that would be the last operation you would perform if you were to actual calculate the function for a particular x value. So the derivative rule for the sine is the first differentiation rule we will use.

So we have the sine of “something” so we start with the derivative of that something: $f'( x) =\text{cos}\left(\sqrt{x^{2} +2x-7} \right)\ ( …)$ Now from the last post, you know you have to multiply this by the derivative of that “something”. It will be helpful to rewrite that “something” as $\sqrt{x^{2} +2x-7} =\left( x^{2} +2x-7\right)^{1/2}$ Looks like we need to apply the chain rule again as we have an inner (polynomial) and an outer (power) functions.

The derivative of the “something” to the 1/2 power is $\frac{1}{2}\left( x^{2} +2x-7\right)^{-1/2}( …)$

We are now left with the innermost function x² + 2x – 7. The chain rule says to multiply the previous results with the derivative of this innermost function which is 2x +2. So putting this in the last (…) and then putting that result in the first (…) gives $f'( x) =\frac{1}{2}\left( x^{2} +2x-7\right)^{-1/2}( 2x+2)\text{cos}\left(\sqrt{x^{2} +2x-7}\right)$ Do you see how the successive differentiations of the functions from the outermost to the innermost works with the chain rule?

Let’s do another example. Let’s differentiate $f( x) =\sqrt{\text{sin}( x)\text{cos}( x)}$

As we did before, it’s easier to see the applicable differentiation rule if you convert the square root to its equivalent exponent form:$f( x) =\left[{\text{sin}( x)\text{cos}( x)}\right]^{1/2}$

Hopefully you can now identify the outermost operation as raising “something” to the 1/2 power. So the power rule is the one to use at first:$f'( x) =\frac{1}{2}[\text{sin}( x)\text{cos}( x)]^{-1/2}( …)$

So we now need to multiply this by the derivative of the “something” which is sin(x)cos(x). But this is the multiplication of two functions so we need to use the multiplication rule. Letting u = sin(x) and v = cos(x), then uv + uv‘ becomes cos²(x) – sin²(x). So now replacing the (…) with this results in $f'( x) =\frac{1}{2}[\text{sin}( x)\text{cos}( x)]^{-1/2}\left[\text{cos}^{2}( x) -\text{sin}^{2}( x)\right]$

This last example highlights the point that to find the derivative of complex functions frequently requires the use of several differentiation rules. You need to be aware of where you are in a particular problem and which rule you are currently working on.

Next time, I will show some examples where the derivatives are used. In the meantime, you can use the results of derivatives found in this post to find the derivative of$f( x) =\sqrt{\text{sin}( x)\text{cos}( x)}\text{sin}\left(\sqrt{x^{2} +2x-7}\right)$