Another Trig Diff Example

[Jason76]

Product Rule:

Given: \(\displaystyle f(x) g(x)\)

\(\displaystyle g(x)[f'(x)] + f(x)[g'(x)]\)

\(\displaystyle f(x) = \sec(x)\tan(x)\)

\(\displaystyle f'(x) = \tan(x)[\dfrac{d}{dx} sec(x)]+ \sec(x)[\dfrac{d}{dx} \tan(x)]\)

\(\displaystyle f'(x) = \tan(x)[\sec(x)\tan(x)] + \sec(x)[sec^{2}(x)]]\)

\(\displaystyle \sec(x)\tan(x)\tan^{2}(x) + \sec^{3}{x}\)

 

[lookagain]

Product Rule:

Given: \(\displaystyle f(x) g(x)\)

\(\displaystyle g(x)[f'(x)] + f(x)[g'(x)]\)

\(\displaystyle f(x) = \sec(x)\tan(x)\)

\(\displaystyle f'(x) = \tan(x)[\dfrac{d}{dx} sec(x)]+ \sec(x)[\dfrac{d}{dx} \tan(x)]\)

\(\displaystyle f'(x) = \tan(x)[\sec(x)\tan(x)] + \sec(x)[sec^{2}(x)]]\)

\(\displaystyle \sec(x)\tan(x)\tan^{2}(x) + \sec^{3}{x}\) <-------

\(\displaystyle \ \ \ \)No, you have one too many tan(x) factors to the left of the plus sign.



\(\displaystyle \sec(x)\tan^2(x) \ + \ \sec^3(x) \ \ \ \ \ \) This is what it should be up to this point.

What might you do next?

Edit: You can't use "f(x)" to refer to the original function, and to one of the lesser functions that makes up a factor of the function, at the same time in the problem.