How do I differentiate "messy" functions?

 

Often, it can seem daunting when you are asked to differentiate a function that looks particularly messy. However, it's really quite simple if you break it down in easy to manage steps.

I find students seem to hate square roots in particular. So let's start with a worked example and then i'll provide a general outline of how to deal with what I call "compound" functions. Before you look at this, you should be aware of the product and chain rules of differentiation. 

1) Easy example: Differentiate x^1/2. 

You can use your standard rule of "multiply by the power and take off 1" here. The result is 1/2x^-1/2. 

2) Tougher example: Differentiate (log x)^1/2. 

Okay. This looks a bit more challenging! However...we can use substitution and the chain rule to get us through this.

Chain rule: If we have functions u(x) and v(x), we can swiftly differentiate the composition of the two functions u(v(x)) by the following rule.
u(v(x))' = u'(v(x))*v'(x), where u' refers to d/dx u(x), and likewise for v. 

This is useful! Now we can use our result in part 1 again. We substitute log(x) for v(x) and x^1/2 for u(x), then our function (log x)^1/2 can be seen as the composition of x -> v(x) = log(x) ->u(v(x)) = (log x)^1/2, and it reads as v^1/2 instead. Differentiating this, as if "v" were "x," we get 1/2v^-1/2. This part is u'(v(x)) in the chain rule, so we are not done yet! We still need to multiply by v'(x)!
Hopefully, you know that differentiating log x gives x^-1. 
Putting this all together, we now know ((log x)^1/2)' is 1/2(log x)^-1/2x^-1. 

3) Hard example: Differentiate (3xsin(x))^1/2.

Learning from the above, we can do something similar. If we substitute 
v(x) = 3xsin(x)
u(x) = x^1/2

Then we get the composition 

x -> v(x) = 3xsin(x) -> u(v(x)) = (3xsin(x))^1/2.

Again remember the chain rule: u(v(x))' = u'(v(x))*v'(x). 

Then our result is 1/2(3xsin(x))^-1/2(3xsin(x))'
So we just need to find out what (3xsin(x))' is.

Remember the product rule? 

Product rule: If we have two functions f(x) and g(x), we can differentiate their product fg(x) by the following rule:
(fg(x))' = f'(g(x)) + g'(f(x)). 

Okay! So here we could view 3x as f(x), and sin(x) as g(x). Then we can find
f'(x) = 3
g'(x) = cos(x) 
and use the formula to get
(3xsin(x))' = 3sin(x) + 3xcos(x) (this is v'(x))

Ok, good! Now we have everything, we just multiply it all together!

((3xsin(x))^1/2)' = 1/2(3xsin(x))^-1/2[3sin(x) + 3xcos(x)] 

which simplifies to 3^1/2(sin(x) + xcos(x)) / (2(xsin(x))^1/2). (I hope you agree!)

 

This technique is valuable when you are given a nasty looking function to differentiate. It can be used multiple times to break the differentiation down into nice "bite-sized" pieces that are easy to manage. 
For example, you could start with something horrid, like
(4x²e^xsin(5x))^1/2 (eek!) and your first step would be to use 
v(x) = 4x²e^xsin(5x)
u(x) = x^1/2 

and use the chain rule like before. You then have to differentiate what's in the bracket using more substitution, and a combination of chain and product rules, many times!

So, to conclude, the method is

1. Look at your function and see what would be a useful substitution (what I called u(x) and v(x) above)
   (I only referred to square roots above, so to help you get the idea, ill give another quick example here: If I had to differentiate log(x³sin(x)) I would start with v(x) = x³sin(x) u(x) = log(x)) 
   
2. You should now have something that is a function of v(x) (the above would be log(v(x)), for example). Figure out which differentiation rule would be best to use here!
   
3. Now, you can differentiate the two pieces of the function separately!
   
   It may be necessary to use the process again on each of the pieces, but if you take care and write down the process clearly, it shouldn't be too much of a problem to multiply it all together at the end!