Question

f''(pi/4)=sec(x)

Answer 1

Is that second derivative of f?

Answer 2

yes

Answer 3

wait... it seems there's a syntax error in what you have there

Answer 4

whaat?>

Answer 5

what you posted does make sense, but f''(x)=sec(x) does if the problem asks you to find f'' at pi/4

Answer 6

does not* make sense - sorry

Answer 7

sorry. was helping some one and typing this. it is What is f''(pi/4) if f(x)=sec(x)?

Answer 8

Ah, that makes more sense!

Answer 9

Do you know how to take the derivative of sec(x)?

Answer 10

yeah
i got to sec^3(x)+tan^2(x)sec(x)

Answer 11

\[\sec(x) = \frac{1}{\cos(x)}\]
Can use the quotient rule on that and get
\[\frac{\sin(x)}{\cos^2(x)} = \sec(x)\tan(x)\]
as the first derivative.

Answer 12

Ok, yeah, you got the second derivative.
Now just plug in π/4 to that.

Answer 13

but my assignment says its wrong
i got -7.992

Answer 14

I don't know how you got -7.992. If you're using a scientific calculator, make sure it's in radian mode.

Answer 15

Might be safer to do it by hand: cos(π/4) = sin(π/4) = √(2)/2.

Answer 16

now i get one but it says thats wrong

Answer 17

Hmm, I'm pretty sure this is the correct second derivative:
\[\sec^3(x)+\tan^2(x)\sec(x)\]
\[x=\frac{π}{4} \rightarrow \sec^3(\frac{π}{4})+\tan^2(\frac{π}{4})\sec(\frac{π}{4})\]

Answer 18

\[\cos(\frac{π}{4})=\frac{\sqrt{2}}{2} \rightarrow \sec^3(\frac{π}{4})=\frac{2^3}{\sqrt{2}^3}\]
\[\tan(\frac{π}{4})=1\]

. . .

Answer 19

Should be, then:
\[\frac{2^3}{\sqrt{2}^3}+1 \cdot \frac{2}{\sqrt{2}} = \frac{8}{\sqrt{8}}+\frac{2}{\sqrt{2}} = 3\sqrt{2}.\]

Answer 20

I think.. might want to double check me on all that.