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

##### 2 Answers

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.)

Differentiate

Multiply through b

Solve for

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

Now differentiate again.

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

Use

# = 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

# = 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

#### Explanation:

Differentiate the equation with respect to

Solving for

Differentiating again: