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## roselin Group Title Find the derivative of the function-using the chain rule. k(x)= x^2 sec(1/x) one year ago one year ago

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1. satellite73 Group Title

need the product rule as well

2. satellite73 Group Title

start with $2x\sec(\frac{1}{x})+x^2\frac{d}{dx}\sec(\frac{1}{x})$ second part requires chain rule

3. roselin Group Title

okay

4. roselin Group Title

the one that you have done is by using the product rule?

5. satellite73 Group Title

the derivative of secant is secant tangent, and the derivative of $$\frac{1}{x}$$ is $$-\frac{1}{x^2}$$

6. satellite73 Group Title

yes

7. roselin Group Title

okay,

8. satellite73 Group Title

so the whole thing is $2x\sec(\frac{1}{x}+x^2\sec(\frac{1}{x})\tan(\frac{1}{x})\times (-\frac{1}{x^2})$

9. satellite73 Group Title

we can clean it up a bit as $2x\sec(\frac{1}{x})-\sec(\frac{1}{x})\tan(\frac{1}{x})$

10. satellite73 Group Title

on account of the $$x^2$$ cancel

11. roselin Group Title

okay,

12. roselin Group Title

If anytime i get a problem like this, do I have to use the product rule first and then continue with the chain rule?

13. satellite73 Group Title

well 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

14. roselin Group Title

oh okay

15. satellite73 Group Title

just 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

16. roselin Group Title

i have one more question/problem.would you be willing to help me out ?