Find y'' by implicit differentiation. 7x^2 + y^2 = 6

Question

Calculus I Question. Need to find y'' for the following expression.

Answer 1

7x² + y² = 6

 

Differentiate both sides.

 

14x + 2yy' = 0

 

y' = -14x/(2y) = -7x/y

 

Differentiate both sides.

 

y'' = -7(y - xy')/y²

 

Plug in y' from above.

 

y'' = -7[y - x(-7x/y)]/y² = -7(y + 7x²/y)/y² = -7[(y² + 7x²)/y]/y³ = -7(y² + 7x²)/y³

 

Note this is the same as Thomas' answer, but it can be simplified a little more.

 

y² + 7x² = 6 ⇒

 

y'' = -7(6)/y³ = -42/y³

 

I do not see how this can be simplified further.

Answer 2

1st derivative: 14x +2y(dy/dx) = 0  solve for dy/dx

 

dy/dx = -7x/y

 

Second derivative: d²y/dx² = [-7(y) - 1(dy/dx)(-7x)]/y²    substitute -7x/y for dy/dx and simplify

 

(-7y -49x²/y)/y² = (-7y² -49x²)/(y³)