How do I find the derivative of 9secx-2x?

Question

I need to find the derivative of 9secx-2x.  I am stumped.

Answer 1

Hello, thank you for taking the time to post your question!

You just want to go term by term … the derivative of 9sec(x) is 9sec(x)tan(x) and the derivative of -2x is -2, so putting it all together you get

d/dx(9 sec x – 2x)

= 9sec(x)tan(x) – 2

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

Let f(x) = 9sec(x)

Let g(x) = 2x

 

So the problem is f(x) - g(x)

 

The derivative of a difference of two functions is just:

 

(f - g)' = f ' - g'

 

Now f ' = d(9sec(x))/dx = 9*dsec(x)/dx = 9sec(x)tan(x)     [See note below]

 

g' = d(2x)/dx = 2*dx/dx = 2

 

So (f-g)' = f ' - g' = 9sec(x)tan(x) - 2

 

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Note:

 

sec(x) = 1/cos(x) = cos(x)^-1

 

d(cos(x)^-1)/dx = -cos(x)^-2*(-sin(x)) = sin(x)/cos²(x) = sin(x)/cos(x)cos(x) = tan(x)sec(x)