Question

If f(x) = tan^(2)x + sinx, then f'(pi/4) =

Answer 1

Do you know how to find f'(x)?

Answer 2

would it be \[f'(x)=2tanxsec^2x+cosx\] ?

Answer 3

would it? :o

Answer 4

that's right

Answer 5

Okay i just have issues with the rest when evaluating f'(pi/4)

Answer 6

What are tan(pi/4) and cos (pi/4)?

Answer 7

tan(pi/4)=1 and
cos(pi/4)=\[\frac{ 1 }{ \sqrt{2} }\]

Answer 8

Nice :)
What is sec(pi/4)? Note that sec(theta) = 1/[cos(theta)]

Answer 9

so that would equal to 2?

Answer 10

sec(pi/4) = 1/cos(pi/4) = \(\large \frac{1}{\frac{1}{\sqrt{2}}}\) = \(\sqrt{2}\)
So, sec^2 (pi/4) = 2.
Does that make sense to you?

Answer 11

yes i understand that, so now how would the final answer look put together

Answer 12

f'(x) = 2tanx sec^2(x) + cosx
f'(pi/4) = 2 tan(pi/4) sec^2(pi/4) + cos(pi/4)
Just now, you have evaluated tan(pi/4), sec^2(pi/4) and cos(pi/4), just substitute the values you got into f'(pi/4), so you get
\[f'(\frac{\pi}{4}) = 2 tan(\frac{\pi}{4}) sec^2(\frac{\pi}{4}) + cos(\frac{\pi}{4})\]
\[f'(\frac{\pi}{4}) = 2(1)(2)+\frac{1}{\sqrt{2}}=...???\]

Answer 13

would it be \[\frac{ 4\sqrt{2}+1 }{\sqrt{2} }\]

Answer 14

Yeah! @ifunfrank it is \(\cfrac { 4\sqrt{2} + 1 }{\sqrt{2} }\)

Good work!

Answer 15

You can also rationalize it : \(\cfrac{4\sqrt{2}+1}{\sqrt{2}} \times \cfrac{\sqrt{2}}{\sqrt{2}}\)
\(\implies \cfrac{8 + \sqrt{2}}{2}\)