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 f(x) = g(x)+h(x)\displaystyle f(x) \ = \ g(x) + h(x)f(x) = g(x)+h(x) then, df(x)dx = dg(x)dx+dh(x)dx\displaystyle \frac{df(x)}{dx} \ = \ \frac{dg(x)}{dx} + \frac{dh(x)}{dx}dxdf(x) = dxdg(x)+dxdh(x) 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 f(x) = g(x)+h(x)\displaystyle f(x) \ = \ g(x) + h(x)f(x) = g(x)+h(x) then, df(x)dx = dg(x)dx+dh(x)dx\displaystyle \frac{df(x)}{dx} \ = \ \frac{dg(x)}{dx} + \frac{dh(x)}{dx}dxdf(x) = dxdg(x)+dxdh(x) 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. ddx[−x + tan(x)] = −1 + sec2(x)\displaystyle \frac{d}{dx}[-x \ + \ tan(x)] \ = \ -1 \ + \ sec^2(x)dxd[−x + tan(x)] = −1 + sec2(x) You could further simplyfy it using: sec2(x) = 1 + tan2(x)\displaystyle sec^2(x) \ = \ 1 \ + \ tan^2(x)sec2(x) = 1 + tan2(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. ddx[−x + tan(x)] = −1 + sec2(x)\displaystyle \frac{d}{dx}[-x \ + \ tan(x)] \ = \ -1 \ + \ sec^2(x)dxd[−x + tan(x)] = −1 + sec2(x) You could further simplyfy it using: sec2(x) = 1 + tan2(x)\displaystyle sec^2(x) \ = \ 1 \ + \ tan^2(x)sec2(x) = 1 + tan2(x)
