Alright, if anyone in here has even just been in elementary calculus, they'll know what I'm talking about. I've missed too many days of school (and therefore my Calculus classes) and I need some help with the chain rule of derivatives.

Is there anyone here that has the capability to explain to me, quite simply, what I should do with the chain rule, and how to apply it with some examples (mix and match rules if you want, let's have a trig function party!)?

Or would it be better for me to try and decipher it from Wolfram and other websites...?

I'll keep reading, but some help would be nice.

Oh boy, I despise calculus. I will assume that you have a basic understanding of differentiation and know the power rule.

Okay, first we must examine the definition of the chain rule. It states that if y is a differentiable function of u and u is a differentiable function of x, then the derivative of y with respect to x is given by:

dy/dx = dy/du * du/dx

Alternatively, if f(x) = g(h(x)), then f '(x) = g '(h(x)) * h '(x).

So what does that actually mean? It is essentially saying that in complex differentiation, the derivative of the "outside" function comes first, followed by the "inside" function, and you multiply them together. Here's an example:

f(x) = (x^3 + 7)^2 -- The stuff in the parentheses is squared, so that will be your outermost function.
......= 2(x^3 + 7) -- Use the power rule to bring down the exponent.
......= 2(x^3 + 7) * 3x^2 -- Use the power rule again to find the derivative of the inside and multiply.
f '(x) = 6x^2 * (x^3 + 7) -- Simplify and you are done.

Does that make any sense? Here is another example, with some trigonometry:

g(x) = sin(5x) -- There are no exponents in this one, so sine is the outside function.
......= cos(5x) -- Recall that the derivative of sine is cosine.
......= cos(5x) * 5 -- Differentiate the inside and multiply. The derivative of 5x is simply 5.
g '(x) = 5cos(5x) -- Simplify. That one was easy, right?

I have one more example that combines both trigonometry and exponents:

h(x) = cos^4(2x) -- This one is tricky. You have to realize that cos^4 is the outside function.
......= (cos(2x))^4 -- Recall that cos^4(x) is the same as saying (cos(x))^4.
......= 4(cos(2x))^3 -- Use the power rule to bring down the exponent.
......= 4(cos(2x))^3 * (-sin(x)) -- Take the derivative of cosine and multiply.
......= 4(cos(2x))^3 *(-sin(x)) * 2 -- Take the derivative of the innermost function, 2x.
h '(x) = -8cos^3(2x) * sin(x) -- Simplify once more to get a clean answer.

I know that was probably confusing, but it is difficult to clearly illustrate the problems over a computer. If you need any more help, send me a PM and I will try to explain it a little better.

I'd give you a hug, but my arms don't reach that far. Thanks a bunch.

I am glad I could be of assistance.