Can someone please help with the below question?

Calculus
Tutor: None Selected Time limit: 1 Day

dy/dt for ln x =e^y+(x^2+y^3)^2-2 if x=1 y=0  and dx/dt=-2

Jul 15th, 2015

Hi, I am pleased to do this question for you. I am giving the detailed solution below.

ln x =e^y+(x^2+y^3)^2-2

differentiating this with respect to dt

d(ln x)/dt = d(e^y)/dt + d((x^2+y^3)^2) - d(2)/dt

(1/x)dx/dt = (e^y)dy/dt + 2*((x^2+y^3)(2x*dx/dt + (3*y^2)dy/dt) - 0      - >  (equation 1)

since

differential of ln x is 1/x 

differential of e^y is e^y but we are doing it with respect to dt so dy/dt term will come

similarly for the next term and differential of constant(2) is always 0.

putting the given values in equation 1,

(1/1)(-2) = (e^0)dy/dt + 2*((1^2 + 0)(2*1*-2 + 0))

-2 = dy/dt + 2*-4

dy/dt = 6      --------->  ANSWER

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 15th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Jul 15th, 2015
...
Jul 15th, 2015
Dec 3rd, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer