Find 𝑑^2𝑥/𝑑𝑥^2 by implicit differentiation y ^ 2 + x ^ 2 = e ^ sqrt(sin(xy) + ln(xy))

Question

Find 𝑑²𝑥/𝑑𝑥² by implicit differentiation y²+ x² = e ^{sqrt(sin(xy) + ln(xy))}

Answer 1

Hello, Haile E,

Problems of that type test not only your mathematical thinking but attention to details in prolonged calculations. That's why it is useful to break it down to some "blocks" and then build the result from such blocks. Below B1 means 1st block etc.

x²+y²=exp{Sqrt[sin(xy)+ln(xy)]} (1)

B1. Simplify (1). The r.h.s. of (1) makes sense when ln(xy)+sin(xy)>=0 or xy>=a where

ln(xy)+sin(xy)=0 [a≅0.579]. For such values of xy , (1) is equivalent to

ln[(x²+y²)]²=sin(xy)+ln(xy) (2)

Let's introduce two functions

t(x,y)=(x²+y²) ; z(x,y)=xy (3)

Then (2) would read as

(ln t )²=sin z + ln z (4)

B2. Differentiate t(x,y)=(x²+y²) , z(x,y)=xy and express dt/dz through x,y, y^'=dy(x)/dx

dt=2xdx + 2yy^'dx=(2x+2yy^')dx ; dz=ydx+xy^'dx=(y+xy^') dx (5a,b)

From (5a,b)

dt/dz=(2x+2yy^')/(y+xy^') (5c)

B3. Using (5c), Express y^' as a function of x,y, dt/dz=t^':

y^'=(2x-yt^')/(2y-xt^') (6)

B4. Express t^'=dt/dz from (4):

d/dt[(ln t )²]=d/dt[sin z + ln z] (7)

[(2ln t)/t] t^'=cos z +1/z or t^'=(cos z +1/z)/[(2ln t)/t]=t(cos(z)+1/z)/2ln t (8)

Finally,

B5. Substitute t^' from (8) in (6):

y^'(x,y)=[ 2x-t(cos(z)+1/z)/2ln t]/[2y-xt(cos(z)+1/z)/2ln t] (9) where t(x,y)=(x²+y²) ; z(x,y)=xy

As to the 2nd derivative of y(x) , you could start with the expression for y^'=y^'[x,y,t(x,y),z(x,y)] in (9)

and differentiate it to find

y^''(x,y)=[x,y,t(x,y),z(x,y), y^'(x,y)] (10)

where y^'(x,y) in (10) is meant to be replaced from (9)

The resulting expression is to become longer than (9).

A note. if you express your result by just differentiating y^'(x,y) from (9) and leave y^' in the r.h.s of (10) "as is" instead of replacing it with its full expression (9) , that might be enough

I hope this would be helpful

Best,

Dr. Ariel B.

Answer 2

Take d/dx in place (using chain rule for y(x)

2yy' + 2x = e^(sqrt(sin(xy)+ln(xy)))*(1/2)*(sin(xy) + ln(xy))^-1/2 (cos(xy)(xy'+y) + (xy'+y)/(xy))

For the e to the power, you can sub in x²+ y²

Group y' terms and solve for y'

Take d/dx again. It''s a mess.