What is the derivative of f(x)= -x+tanx?
T taylor13 New member Joined Dec 5, 2010 Messages 2 Dec 5, 2010 #1 What is the derivative of f(x)= -x+tanx?
D Deleted member 4993 Guest Dec 5, 2010 #2 taylor13 said: What is the derivative of f(x)= -x+tanx? Click to expand... What do you think? If \(\displaystyle f(x) \ = \ g(x) + h(x)\) then, \(\displaystyle \frac{df(x)}{dx} \ = \ \frac{dg(x)}{dx} + \frac{dh(x)}{dx}\) Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.
taylor13 said: What is the derivative of f(x)= -x+tanx? Click to expand... What do you think? If \(\displaystyle f(x) \ = \ g(x) + h(x)\) then, \(\displaystyle \frac{df(x)}{dx} \ = \ \frac{dg(x)}{dx} + \frac{dh(x)}{dx}\) Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.
T taylor13 New member Joined Dec 5, 2010 Messages 2 Dec 5, 2010 #3 the derivative of tanx=sec^2x and i think the derivative of -x=-1 but after that i am lost!
D Deleted member 4993 Guest Dec 5, 2010 #4 taylor13 said: the derivative of tanx=sec^2x and i think the derivative of -x=-1 but after that i am lost! Click to expand... Like I explained - add those up. \(\displaystyle \frac{d}{dx}[-x \ + \ tan(x)] \ = \ -1 \ + \ sec^2(x)\) You could further simplyfy it using: \(\displaystyle sec^2(x) \ = \ 1 \ + \ tan^2(x)\)
taylor13 said: the derivative of tanx=sec^2x and i think the derivative of -x=-1 but after that i am lost! Click to expand... Like I explained - add those up. \(\displaystyle \frac{d}{dx}[-x \ + \ tan(x)] \ = \ -1 \ + \ sec^2(x)\) You could further simplyfy it using: \(\displaystyle sec^2(x) \ = \ 1 \ + \ tan^2(x)\)