Find the differentiation of x4+ y4= 0.
Answers
Answer:
Find y′′ if x4+y4=16 by implicit differentiation
So after the first implicit differentiation I got this equation (let's call it A):
4x3+4y3∗dydx=0 Where dydx is y′
At this point the text book finds the second derivative by making dydx the subject and getting a value of dydx in terms of y and x which is −x3y3 and then taking ddx of −x3y3 to work out d2ydx2. I understand that there is nothing wrong with this approach. However I'm wondering why we can't do implicit differentiation again on Equation A(written above) just like we did on the original equation, in order to find y′′. My reasoning is given below, please tell me what's wrong with it.
ddx(4x3+4y3∗dydx=0)
12x2+[12y2dydx+d2ydx24x3]=0
Making d2ydx2 the subject I get:
d2ydx2=y′′=3x2(x+y)y4
However the textbook gives the answer as:
−3x2(x4+y4)y7 and since x4+y4=16
−3x2(16)y7 = −48x2y7