Question

EXAM HELP PLEASE.

1) Find the slope and y-intercept of the tangent line to the curve f(x) = x^4 + 2e^x when c=0.

2) The derivative of x/(2 - tan x) = [-tan x + xsec(^2)x + 2]/(2 - tan x)^2 OR /(tan x - 2)^2 ???? I would think it's the first one, but the answer WolframAlpha.com gave me has the latter denominator, without showing any last step to get to that.

Answer 1

for the first problem: f'(x) = 4x^3 + 2e^x, right? at least?

Answer 2

then f'(c) = f'(0) = 0 + 2(1) = 2 .....

Answer 3

@calyne you are not making any sense
there is no c in the function

Answer 4

but your derivative is right
now if you mean when x=0, then f'(0)=2, yes

Answer 5

THAT'S ALL THE EXAM SAYS. and isn't the point where the tangent line touches the curve at (c,f(c))??

Answer 6

differentiate the function

dy/dx = 4x^3 + 2e^x

substitute either x = c or x = 0 into your derivate, this gives the gradient at that point

if x = 0 then the gradient m = 2

subtitiute x = 0 into the original equation to find the y value y = 2

so the gradient m = 2 and the point is (0,2) find the equation of the tangent using the point slope formula

Answer 7

so the gradient is m = 4c^3 + 2e^c

Answer 8

alright so the slope is 2, the y-intercept is 2, the equation of the tangent line would be y-2=2(x-0) => y = 2x+2, correct?

Answer 9

@calyne
note: 'c' is just a way of saying that some value of x has been chosen
for example if you write
"if f(x)=x , what is f(x) when c=0", then answer would still be x
if you say "what is f(c)?", then the answer would be "f(c)=c"

Answer 10

listen. i know that. i understand that entirely. okay. i said f(x) before, just to get through the derivative of f. of whatever. that being x. and later, c. which is f(c). which is f(x). or f'(........)

Answer 11

anyway so what about my second question what's the deal there

Answer 12

well let's see
here's a tip: use the product rule and chain rule, \(not\) the quotient rule

Answer 13

and thanks for helping me through this anyway guys i'm in dire need of the emotional support your time is extremely appreciated whether or not it seems like it, i'm just really AGGRAVATED and easily right now it's nothing personal

Answer 14

uh i would but we're specifically instructed that we're NOT up to the chain rule yet, and this is the exam. explicitly preceding the chain rule in my class.

Answer 15

so there's gotta be a way around that

Answer 16

i have the answer all the way down... until that very very last switcheroo step, which is a total mystery

Answer 17

so just show me where that comes from

Answer 18

\[D_x {x\over2 - \tan x}=D_xx(2-\tan x)^{-1}\]\[=(2-\tan x)^{-1}-x(2-\tan x)^{-2}(-\sec^2x)\]\[=(2-\tan x)^{-2}(2-\tan x+x\sec^2x)\]and now a trig identity...

Answer 19

oh fooey, nevermind I didn't read properly
the only problem you have is that you did not remember that\[(a-b)^2=(b-a)^2\]so your answers are the same lol
(extra work for nothing)

Answer 20

yes, please

Answer 21

so\[{2-\tan x+x\sec^2x\over(\tan x-2)^2}={2-\tan x+x\sec^2x\over(2-\tan x)^2}\]that's all ;)

Answer 22

do you understand?

Answer 23

@calyne still with me?

Answer 24

yeah i got it thank you