How do you determine #(d^2y)/(dx^2)# given #x^(2/3)+y^(2/3)=8#?

2 Answers
Jan 6, 2017

Implicit differentiation, stubbornness and attention to details.

Explanation:

The more challenging (difficult) a problem is, the MORE you should write. (As a student I tried to skip steps and it took me a year of two to learn this important fact. If you can answer in two lines, do it in one. If you think it will take 8 lines, plan to write 12.)

#x^(2/3)+y^(2/3) = 8#

Differentiate

#2/3x^(-1/3) + 2/3y^(-1/3) dy/dx =0#

Multiply through b #3/2# to get rid of those fractions.

#x^(-1/3) + y^(-1/3) dy/dx =0#.

Solve for #dy/dx#

#y^(-1/3) dy/dx = - x^(-1/3)#

#dy/dx = -x^(-1/3)/y^(-1/3)#

It's usually silly to have a negative exponent in a fraction, so let's fix that.

#dy/dx = - y^(1/3)/x^(1/3)#

Now differentiate again.

#(d^2y)/dx^2 = ((-1/3 y^ (-2/3) dy/dx)(x^(1/3))-(-y^(1/3))(1/3x^(-2/3)))/x^(2/3)#

# = (- 1/3(x^(1/3)/y^(2/3))dy/dx + 1/3(y^(1/3)/x^(2/3)))/x^(2/3)#

Let's factor out that #1/3# (We could take a #-1/3#, but #dy/dx# has a minus in front of it.)

# =1/3 (- (x^(1/3)/y^(2/3))dy/dx + (y^(1/3)/x^(2/3)))/x^(2/3)#

Use #dy/dx = - y^(1/3)/x^(1/3)# to get

# = 1/3 (- (x^(1/3)/y^(2/3))(- y^(1/3)/x^(1/3)) + (y^(1/3)/x^(2/3)))/x^(2/3)#

# = 1/3 ( (1/y^(1/3)) + (y^(1/3)/x^(2/3)))/x^(2/3)#

Clear the denominators in the numerator by multiplying the big fraction by #((x^(2/3)y^(1/3)))/((x^(2/3)y^(1/3)))#

# = 1/3 (((1/y^(1/3)) + (y^(1/3)/x^(2/3))))/(x^(2/3)) * ((x^(2/3)y^(1/3)))/((x^(2/3)y^(1/3)))#

# = 1/3 (x^(2/3)+y^(2/3))/( x^(4/3)y^(1/3))#

Finally, notice that the numerator is the expression we started with and is equal to #8#.

#(d^2y)/dx^2 = 8/(3x^(4/3)y^(1/3))#

Jan 6, 2017

#y'' =1/3y^(1/3)x^(-4/3)-1/3x^(-1/3)y^(-2/3)y'#

Explanation:

Differentiate the equation with respect to #x#:

#d/(dx) (x^(2/3) + y^(2/3)) = 0#

#2/3x^(-1/3) + 2/3y^(-1/3) y' = 0#

Solving for #y'# we have:

#y'=-x^(-1/3)/y^(-1/3) = -y^(1/3)/x^(1/3)#

Differentiating again:

#y'' = -(1/3x^(1/3)y^(-2/3)y'-1/3y^(1/3)x^(-2/3))/x^(2/3)=1/3y^(1/3)x^(-4/3)-1/3x^(-1/3)y^(-2/3)y'#