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.

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

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

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

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

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

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

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

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?

@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"

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'(........)

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

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

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

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.

so there's gotta be a way around that

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

so just show me where that comes from

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

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)

yes, please

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 ;)

do you understand?

@calyne still with me?

yeah i got it thank you

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