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 Pre-Calculus and Calculus
 derivatives of trig
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effort
Senior Member

USA
38 Posts

Posted - 10/17/2013 :  00:25:13  Show Profile  Reply with Quote
Find the derivative of sec(x).
my work: (d/dx)sec(x)=sec(x)tan (x)(2x)


My question is why is it that there should be only one 2x. According to the chain rule, I suppose to take the derivative of the inside function. Well, there are two x, so why do I not have to write 2x twice. i don't understand.


Also how do you know which is the interior and exterior functions when finding trig derivatives? For example: find dy/dx of y = 4sin(3x)




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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 10/18/2013 :  17:24:09  Show Profile  Reply with Quote
When differentiating sec(x), you get sec(x)tan(x). Note that the arguements are both the same. When you apply the chain rule you apply it to the arguement of the secant (i.e. the orginal function), so you only do it once.
When you start applying the chain rule, you start with the outer part and work your way 'inside'. In the example you used the sine is the outer and the 3x is the inner.
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