Question

what is f'(x) = [2x^2tan(x)] / [sec(x)] ?

Answer 1

i know you have to use the quotient rule and so far I got f'(x) = (4x)(tan(x)) - (sec^2(x))(2x^2) / [sec^2(x)]

Answer 2

but how do you simplify this?

Answer 3

Is it \(f(x) = \frac{2x^2 \cdot \tan(x)}{\sec(x)}\)?
and you want to find \(f'(x)\) now?

Answer 4

yeah

Answer 5

Ok, well in that case, what is \(\sec(x)\)?

Answer 6

1/cos(x)

Answer 7

Right, so we can say:
$$
f(x) = \frac{2x^2 \tan(x)}{\sec(x)} = = \frac{2x^2 \tan(x)}{\frac{1}{\cos(x)}}
$$and we can multiply both the numerator and denominator by \(\cos(x)\) and get:
$$
f(x) = \frac{2x^2 \cdot \tan(x) \cdot \cos(x)}{\frac{1}{\cos(x)} \cdot \cos(x)} = 2x^2 \cdot \tan(x) \cdot \cos(x)
$$

Answer 8

I wrote accidently == instead of =
Now, what is \(\tan(x)\)?

Answer 9

sin(x)/cos(x)

Answer 10

Right, so again:
$$
f(x) = 2x^2 \cdot \tan(x) \cdot \cos(x) = 2x^2 \cdot \frac{\sin(x)}{\cos(x)} \cdot \cos(x) = 2x^2 \cdot \sin(x)
$$

Answer 11

Now it should be much easier to take the derivative using the product rule

Answer 12

So what do you get?

Answer 13

yeah i got 4xsin(x) + 2x^2cos(x) and then i simplified it to 2x(2sin(x) + xcos(x))

Answer 14

exactly =)
http://www.wolframalpha.com/input/?i=+%5B2x%5E2tan%28x%29+%2F+sec%28x%29%5D+%27

Answer 15

okay wow thank you so much! when I first saw this problem I was like how in the world do you solve this

Answer 16

lol, you're welcome =)

Answer 17

so in order to find f'(5) do you just plug in 5 to final answer?

Answer 18

For \(x\) yes

Answer 19

That is the power of derivative, instead of finding a slope in a single point you generate a function for all points

Answer 20

Also, in worst case, that you can't simplify it like we did here, try to do it in parts. First call the numerator \(u\) or something, and use the quotient rule and then calculate \(u'\) separately and plug in wherever you had \(u'\) in the first step.
So in our case we could say:
$$
f(x) = \frac{2x^2 \cdot \tan(x)}{\sec(x)} \\
u = 2x^2 \cdot \tan(x) \qquad u'=(4x \cdot \tan(x)) + (2x^2 \cdot \sec^2(x)) \\
f(x) = \frac{u}{\sec(x)}\\
f'(x) = \frac{(u' \cdot \sec(x)) - (u \cdot \tan(x) \cdot \sec(x))}{\sec^2(x)}
$$And now we can plug in
$$
f'(x) = \frac{(4x \cdot \tan(x) + 2x^2 \cdot \sec^2(x)) \cdot \sec(x) - 2x^2 \cdot \tan(x) \cdot \tan(x) \cdot \sec(x)}{\sec^2(x)}
$$And simplify.. which in our case we could from the beginning.. This is of course very ugly.. but if you have no choice then this is the approach I'd take.

Answer 21

i don't think i'm getting the right answer. i keep getting this weird decimal .0177789598

Answer 22

Well \(\sin(5)\) and \(\cos(5)\) will not give you nice values.. Are you sure you're expecting one for \(x=5\)?

Answer 23

well the question says what is f'(5) using the f(x) that I wrote before. it's part of the same question. first they tell you to find f'(x) and then f'(5)

Answer 24

when i plugged my answer into the computer it marked it wrong

Answer 25

http://www.wolframalpha.com/input/?i=derivative+%282x%5E2+*+tan%28x%29%29%2Fsec%28x%29%2C+x%3D5

Answer 26

The result it gives here is -4.99538 not 0.0177...

Answer 27

i see

Answer 28

make sure that when you calculate sin and cos you do it by radians

Answer 29

ohhh i had it in degree

Answer 30

the derivative \((\sin(x))' = \cos(x)\) is defined that way only when you're working in radians. It can be shown from the definition of the derivative, but if you didn't get there yet then just remember it =)

Answer 31

i'm still not getting in right. pretty sure i'm plugging it wrong into my calculator though

Answer 32

*it

Answer 33

did you try rounding to -5?

Answer 34

oh no my number's not even close to 5. you do plug in 5 for x in this equation right? : f′(x)=(4x⋅tan(x)+2x^2⋅sec2^(x))⋅sec(x)−2x^2⋅tan(x)⋅tan(x)⋅sec(x)sec^2(x)

Answer 35

why are you using this terrible form? use the short one we found
$$f'(x) = 2x(2\sin(x) + x\cos(x))$$
if you plug \(x=5\) this becomes
$$f'(5) = 2 \cdot 5(2\sin(5) + 5\cos(5)) = 20 \sin(5) + 50 \cos(5) = ?$$

Answer 36

oh wow i'm dumb. yeah you get the right answer this way

Answer 37

thanks so much once again!

Answer 38

sure no problem =)

Answer 39

so are you always supposed to simplify as much as you can before finding the derivative to make your life easier or did that only work in this case?