How can i find f''(x) of the function f(x)= 2(cosx)(sec^2x)
I would first simplify it.
f(x)= 2(cosx)(sec^2x)
= 2cosx(1/cos^2x) = 2/cosx = 2secx
f'(x) = 2tanxsecx
y = 2 cos x sec^2x
y' = 2[ cos x d/dx(sec^2x) + sec^2x (-sinx)]
= 2[ cos x{ sec x d/dx sec x +sec xd/dx sec x} -sin x sec^2 x]
= 2 [cos x {2 sec x d/dx sec x} -sin x sec^2 x ]
(but cos x sec x = 1)
= 2 [2 sec x tan x -sin x sec^2 x]
= 2 [2 sin x/(cos^2 x) -sin x/cos^2 x]
= 2 sin x/cos^2 x = 2 tan x sec x
LOL, well more fun to do it the hard way.
LOL, what else to do on a nice warm Sunday afternoon.
Damon, btw, can you take a peak at
http://www.jiskha.com/display.cgi?id=1270812113
Can you see a better way?
I do not see a better way.
To find the second derivative, f''(x), of the function f(x) = 2(cosx)(sec^2x), you can follow these steps:
Step 1: Start by finding the first derivative, f'(x), of the given function. The first derivative can be found using the product rule and the chain rule.
The product rule states that for two functions u(x) and v(x), the derivative of their product is given by (u'v + uv').
The chain rule states that for a composite function g(f(x)), the derivative is given by (g'(f(x)) * f'(x)).
So, applying the product rule and the chain rule, we can find the first derivative of f(x):
f'(x) = (2 * (-sinx) * sec^2x) + (2 * cosx * (2secx * tanx))
Simplifying further:
f'(x) = -2sinxsec^2x + 4cosxsecxtanx
Step 2: Once you have the first derivative, we can differentiate f'(x) again to find the second derivative, f''(x).
Differentiating f'(x) can be done in a similar way as before, using the product rule and the chain rule:
f''(x) = (-2cosxsec^2x - 4sinxsecxtanx) + (-2sinx(2secx)(tanx) + 4cosxsec^2xsec^2x)
Simplifying further:
f''(x) = -2cosxsec^2x - 4sinxsecxtanx - 4sinxsec^4x + 4cosxsec^4x
Combining like terms:
f''(x) = -2cosxsec^2x - 4sinxsec^4x + 4cosxsec^4x - 4sinxsecxtanx
Therefore, the second derivative of f(x) = 2(cosx)(sec^2x) is:
f''(x) = -2cosxsec^2x - 4sinxsec^4x + 4cosxsec^4x - 4sinxsecxtanx