Question

How do you find `(d^2y)/(dx^2)` for `5x+3y^2=1`?

Answer 1

Use product rule, chain rule and sum rule.

Explanation:

Using the sum rule and chain rule to differentiate both sides with respect to `x` gives:

`5+6y*{dy}/{dx}=0`

Differentiating again using the product rule (chain rule is applied in the first term, hence the squared derivative):

`6*({dy}/{dx})^2+6y*{d^2y}/{dx^2}=0`

Rearranging the equation for the first derivative:

`6y*{dy}/{dx}=-5`

`{dy}/{dx}=-5/{6y}`

Substituting this into the equation for the second derivative:

`150/{36y^2}+6y*{d^2y}/{dx^2}=0`

`6y*{d^2y}/{dx^2}=-150/{36y^2}`

`{d^2y}/{dx^2}=-25/{36y^3}`

There you go. If you want it to be in terms of `x`, then you'll end up with some `+-`'s in there. Of course, when `y=0`, the derivative is undefined, which makes sense, since the tangent line at that point is vertical, with infinite slope.