Confused About Which Derivative Rule to Use First

rayroshi

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When trying to find the derivative of (-3x^(3)-1)(csc(5x(4)), I got the wrong answer, because I used the chain rule instead of the product rule first. How am I to know which rule to use first?
 
When trying to find the derivative of (-3x^(3)-1)(csc(5x(4)), I got the wrong answer, because I used the chain rule instead of the product rule first. How am I to know which rule to use first?
You have two factors:
[imath](-3 x^3 - 1)[/imath]

and
[imath]\csc(5x(4))[/imath]
(Whatever you mean by (4).)

So... Product rule.

-Dan
 
When trying to find the derivative of (-3x^(3)-1)(csc(5x(4)), I got the wrong answer, because I used the chain rule instead of the product rule first. How am I to know which rule to use first?
Work from the outside in: Whatever operation is done last in evaluating, apply the appropriate rule.

If the expression is, ultimately, a product, use the product rule. If one of the factors is then a composite function, use the chain rule.
 
Thanks for that! I had forgotten what Mark Ryan said in his book "Calculus for Dummies," (my level) when he advised the same, telling the reader to imagine they are putting the expression to be differentiated into a calculator. He said that the last thing to be entered would be the first thing to be dealt with in the chain of operations. I guess that the whole idea of working from the outside inward just seems to be opposite to the old 'PEMDAS' mnemonic, where you have to do what is in the parentheses first.
 
Thanks for that! I had forgotten what Mark Ryan said in his book "Calculus for Dummies," (my level) when he advised the same, telling the reader to imagine they are putting the expression to be differentiated into a calculator. He said that the last thing to be entered would be the first thing to be dealt with in the chain of operations. I guess that the whole idea of working from the outside inward just seems to be opposite to the old 'PEMDAS' mnemonic, where you have to do what is in the parentheses first.
This happens in many places; for example, in solving a simple equation, you undo each operation, working from the outside in, layer by layer, until you "set free" the variable that is on the inside. Here, the rules you apply each transform a function of something into its derivative in terms of the derivative(s) of the "something", which you then have to deal with. You can actually work from the inside out in a sense, but you don't know what to do with the derivative of the inner function(s) until you've dealt with the outer function.

When I say that, I'm thinking of the product rule as applying to an outer function "product(u, v)" with inner functions u and v, so the derivative is "sum(product(u', v), product(u, v'))". That gives you places to put u' and v', which you may have actually evaluated previously. The same is true, more simply, of the chain rule.

Think of PEMDAS as gravity; sometimes you want to go downhill, and sometimes you want to go uphill.
 
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