2 quiz questions. Had to solve each under 5 mins. I got the answers wrong, but I'd like to know how the bloody hell to get the correct ones. -_- Any help is very much appreciated because I feel like I want to kill myself right now.

Q1) y=1, x=0, Find y' in x^3 + y^3 = (x+y)^2 - 3(x-y)^4 +1

Q2) y"(0) for x= cos(t/1+t), y = sin (t/1+t)

They may be easy, but I'm frankly terrible at calculus. Please help.

Question

2 quiz questions. Had to solve each under 5 mins. I got the answers wrong, but I'd like to know how the bloody hell to get the correct ones. -_- Any help is very much appreciated because I feel like I want to kill myself right now.

Q1) y=1, x=0, Find y' in x^3 + y^3 = (x+y)^2 - 3(x-y)^4 +1

Q2) y"(0) for x= cos(t/1+t), y = sin (t/1+t)

They may be easy, but I'm frankly terrible at calculus. Please help.

gibbs · Answer

sorry inm in seventh grade

anonymous · Answer

$$x^3 + 1^3 = (x+ 1)^2 - (3x - 3)^4 + 1$$

anonymous · Answer

$$x^3 + 1 = (x^2 + 1) - (3x^4 - 81) + 1$$

anonymous · Answer

Sorry, nevermind, I'm so confused.

xapproachesinfinity · Answer

It's just implicit differentiation with respect to x 
what did you do

xapproachesinfinity · Answer

How did you start.

xapproachesinfinity · Answer

I'm on the phone right now so I can really help that much.

anonymous · Answer

I get the rules I need to apply. It's a chain rule question and an implicit diff. question but the issue is getting to the final answer. I want someone to go through it so I can follow through with my steps and see where I went wrong.

xapproachesinfinity · Answer

Alright let's see here
$\large x^3+y^3=(x+y)^2-3(x-y)^4+1$
differentiate both sides you get
$\large 3x^2+3y'y^2=2(1+y')(x+y)-12(1-y')(x-y)^3$
did you do this

xapproachesinfinity · Answer

replace x=0 and y=1 and simplify

xapproachesinfinity · Answer

@GlitchMob  are you there
if you replace x and y by their values it should be easy you will get only one variable y'

xapproachesinfinity · Answer

$\large 3y'=2(1+y')+12(1-y')$
+12... because i have (-1)^3=-1 with - i got +
now it is real;y just a simple equation (linear equation that is)

xapproachesinfinity · Answer

for the part 2 are you looking for y"(0)?

anonymous · Answer

Yup. Also, can you just redo that final step where you got -1? I reached that equation but I didn't get -1

anonymous · Answer

The answer for Q1) is 11/12, and Q2) is -1

xapproachesinfinity · Answer

oh, i had the last term in the right hand side like this $\large -12(1-y')(x-y)^3$
if i replaced x=0 and y=1
i get $\large  -12(1-y')(0-1)^3=-12(1-y')(-1)^3=(-12)(1-y')(-1)=12(1-y')$

xapproachesinfinity · Answer

hmm if you used that it should be good
i didn't go further to see the final answer for the first part