Trigonometry and the Chain RuleDate: 12/01/98 at 17:59:54 From: Amanda Subject: Calculus - the trig unit I have three questions that have me stumped. I need to differentiate the following: y = 2 csc^3(sqrt(x)) y = x/2 - (sin (2x))/4 y = (1 - cos (x))/sin (x) Date: 12/01/98 at 19:37:16 From: Doctor Santu Subject: Re: Calculus - the trig unit Amanda: These all have to do with the Chain Rule. Here's the basic idea. Suppose: y = sin (x^3 + tan (x)). How do you find the derivative? Think of x^3 + tan x as a big BLOB. So we really need to find the derivative of: y = sin (BLOB) Well, the rule says that the derivative of sin (BLOB) is simply cos (BLOB) multiplied by the derivative of the BLOB itself. A word on notation: I'm going to write y' for the derivative of y (instead of dy/dx). Now, in this case: y' = cos (x^3 + tan(x)) * (3x^2 + sec^2 (x)) because BLOB is x^3 + tan (x), and the derivative of the BLOB is 3x^2 + sec^2 (x). Let's try another example: y = sin^3 (x^3 + tan x) This is really: y = [ sin (x^3 + tan x) ]^3 Using our previous terminology, the derivative of (blob)^3 is simply: 3(blob)^2 * (the derivative of blob itself). In this case: y' = 3[sin (x^3 + tan (x))]^2 * (derivative of sin (x^3 + tan(x))) = 3[sin(x^3 + tan (x))]^2 * cos (x^3 + tan (x)) * (derivative of x^3 + tan (x)) = 3[sin(x^3 + tan(x))]^2 * cos(x^3 + tan(x)) * (3x^2 + sec^2(x)) For another example: y = csc^3 (x^3 + tan (x)) It's the same as: y = [csc (x^3 + tan (x))]^3 So: y' = 3[csc(x^3 + tan (x))]^2 * (deriv. of csc (x^3 + tan (x)) ) = 3[csc(x^3 + tan (x))]^2 * (-csc(x^3 + tan (x)) * cot(x^3 + tan (x)) * (x^3 + tan (x))') = 3[csc(x^3 + tan (x))]^2 * (-csc(x^3 + tan (x)) * cot(x^3 + tan (x)) * (3x^2+sec^2 (x))) For example 3, try: y = csc^3( sqrt (x) ) Remember it's simply: y = [csc(sqrt (x))]^3 So it should start off with: 3[csc(sqrt (x))]^2 * (the derivative of the cosecant...) To take the derivative of [sin (2x)]/4, just rewrite it as: (1/4) * sin (2x) The 1/4 is just a constant, so it just sits there, and the derivative would be: (1/4) * cos (2x) * (2) = (1/2) * cos (2x) In the chain rule, the basic idea is to peel the onion from the outside. You want to take the derivative of a function within a function within a function. You take the derivative of the outermost function relative to the stuff that's inside it, then multiply that by the derivative of the inside expression, relative to the expression inside the expression, and so on, all the way down to the tiniest little x all the way inside. (And some people even stick a "1" on at the end, because the derivative of an x is just 1. I think that's overdoing it a bit.) The Chain Rule needs quite a lot of imagination to see these formulas as expressions within expressions, and ideally you should have a friend sit by you and point out how to "peel the onion" layer by layer. One final example: y = sin (tan (sin^2 (x^7 + 3x) ) ) y' = ...? You must first take the derivative of sin (expression), relative to the expression that's inside. You multiply that by the derivative of the tan (inside expression). You multiply that by the derivative of [sin (x^7 + 3x)]^2, because sin^2 (x^7+3x) means [sin(x^7+3x)]^2. That, in turn, will contain the derivative of sin(x^7 + 3x), which in turn will contain the derivative of (x^7 + 3x), which is 7x^6 + 3. It's important to put the proper expression inside the various partial expressions. So: y' = cos(tan ...) * sec^2(sin ...) * 2[sin ...] * cos(x^7 + 3x) * (7x^6 + 3) You have to know what I have left out, and you must know how to put it in. I suggest you complete the derivative of that derivative just above, inserting all the expressions that would take the place of the ...s, then try the problems you're interested in. (All of us at Dr. Math had to practice these too.) - Doctor Santu, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/