If f(x)= sec x, find f"(Pi/4)
I am not sure how to take the 2nd derivative?
f'(x)= sec x tan x
f"(x)=???
Is it f"(x)= (sec x tan x)(sec^2x)???
Please Help!
f'(x)= sec x tan x is correct
now use the product rule
f''(x) = secx(sec^2x) + tanx(secx)(tanx)
= sec^3 x + secx(tan^2x)
I will leave the subbing to you
What do you mean subbing? I thought that was the final answer? In the book it leaves f"(x)= - cos x for f(x)= cos x
Oops, sorry about that. Do you mean this:
sec^3(Pi/4) + sec(Pi/4)(tan^2(Pi/4)) =
3*sqrt(2)
To find the second derivative of f(x) = sec x, you need to differentiate the first derivative, f'(x), again.
First, let's find the first derivative of f(x) = sec x:
f'(x) = d/dx(sec x) = sec x tan x
Now, let's find the second derivative:
f"(x) = d/dx(sec x tan x)
To differentiate this expression, we can use the product rule.
The product rule states that for two functions, u(x) and v(x), the derivative of their product is given by:
d/dx(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
In this case, we have u(x) = sec x and v(x) = tan x.
Let's apply the product rule to find the second derivative:
u'(x) = d/dx(sec x) = sec x tan x
v'(x) = d/dx(tan x) = sec^2 x
Now, apply the product rule:
f"(x) = u'(x)v(x) + u(x)v'(x)
= (sec x tan x)(tan x) + (sec x)(sec^2 x)
Therefore, the second derivative of f(x) = sec x is:
f"(x) = (sec x tan x)(tan x) + (sec x)(sec^2 x)
To find f"(π/4), you can substitute π/4 for x in the expression we just found:
f"(π/4) = (sec(π/4) tan(π/4))(tan(π/4)) + (sec(π/4))(sec^2(π/4))
Evaluating this expression will give you the value of the second derivative at x = π/4.