find the second derivative of f(x)=sec(x)

Question

anonymous · Answer

$$f(x) = \sec(x) = \frac{1}{\cos(x)}=\cos(x)^{-1}$$
Chain rule states that if $$f(x) = g(x)^{n}$$ then $$f'(x) = n*g(x)^{n-1} * g'(x) $$
if we follow that method then, $$f'(x) = -cos(x)^{-2}*(-sin(x))$$
which simplifies to $$f'(x)=sin(x)sec^{2}(x)$$

anonymous · Answer

for the second derivative, we have a combination of chain rule and product rule.
If you recall, product rule works as follows: $$d(x) = g(x)h(x)$$
$$d'(x)=g'(x)h(x) + h'(x)g(x)$$
therefore we can substitute into this equation $$d(x) = sin(x)cos^{-2}(x) =g(x)h(x)$$ where $$sin(x) = g(x)$$ and $$h(x) = cos^{-2}(x)$$
Now we have to find h'(x) and g'(x), plug it back in to find d'(x)
You should also note that $$d'(x) = f''(x)$$ which is your final answer

no_ellephant · Answer

f''(x)=sec(x)(tan^2x + sec^2x)   ???? yes,, no?

anonymous · Answer

no $$g(x) = sin(x)$$
$$g'(x) = cos(x)$$
$$h(x) =cos^{-2}(x)$$
$$h'(x) = 2sin(x)cos^{-3}(x)$$
therefore
$$f''(x)=d'(x)=g(x)h'(x)+g'(x)h(x)=2tan^{2}(x)sec(x) + sec(x)$$

no_ellephant · Answer

actually I figured it out and you ARE wrong. f'(x) = sec(x)tan(x) then http://www.wolframalpha.com/input/?i=sec%28x%29+tan%28x%29+derivative

anonymous · Answer

if you look at my first derivative $$f'(x) = sin(x)sec^{2}(x)$$ you'll notice that you can write that in multiple ways. $$sec^{2}(x) = cos^{-2}(x)$$ which if you substitute into f'(x), you'll get $$f'(x) = sin(x)cos^{-2}(x) = tan(x)cos^{-1}(x) = tan(x)sec(x)$$