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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0need the product rule as well

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0start with \[2x\sec(\frac{1}{x})+x^2\frac{d}{dx}\sec(\frac{1}{x})\] second part requires chain rule

roselin
 2 years ago
Best ResponseYou've already chosen the best response.0the one that you have done is by using the product rule?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0the derivative of secant is secant tangent, and the derivative of \(\frac{1}{x}\) is \(\frac{1}{x^2}\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0so the whole thing is \[2x\sec(\frac{1}{x}+x^2\sec(\frac{1}{x})\tan(\frac{1}{x})\times (\frac{1}{x^2})\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0we can clean it up a bit as \[2x\sec(\frac{1}{x})\sec(\frac{1}{x})\tan(\frac{1}{x})\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0on account of the \(x^2\) cancel

roselin
 2 years ago
Best ResponseYou've already chosen the best response.0If anytime i get a problem like this, do I have to use the product rule first and then continue with the chain rule?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0well it is not really a matter of "what goes first" you have to use the rules as you need them \(x^2\sec(\frac{1}{x})\) is a product so you need the product rule for sure also \(\sec(\frac{1}{x})\) is a composite function, so you must use the chain rule when you take the derivative

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0just like if you have a quotient, you have to use the quotient rule, but if the numerator is a product, you will need the product rule for that one and if the denominator is a composite function you will need the chain rule for it use whatever rules you need to get the derivative

roselin
 2 years ago
Best ResponseYou've already chosen the best response.0i have one more question/problem.would you be willing to help me out ?
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