Find y'' by implicit differentiation.

Question

Find y'' by implicit differentiation.

8x²+y²=5

y''=_______

Answer 1

Hello Simon!

I hope I can help answer your question today.

Implicit differentiation is really fun and cool, but it takes a bit of practice to get used to using it.

So we have

8x² + y² = 5

We are going to differentiate both sides with respect to x.

So we get:

16x¹ + 2y * dy/dx = 0

So, when we take the derivative of y² we have to use the chain rule, because we do not know what y is.

We are taking the derivate of y with respect to x, so we multiply by dy/dx. (as the chain rule dictates).

It might help to realize that we can do the same thing with 8x².

d/dx( 8x²) = 16x * dx/dx.

We can use the chain rule here too, but dx/dx is just 1, so it goes away.

If we had, say, 8(3x² + 1)², instead, however, the chain rule would be much more interesting:

2*8(3x² + 1)^2-1 * d/dx (3x² + 1)

= 16(3x² + 1)¹ * (2*3x^2-1 + 0)

= 16(3x² + 1)¹ * (6x)

Which we could then combine the terms and simplify.

Anyway, getting back to the original problem:

We have:

16x¹ + 2y * dy/dx = 0

We want to solve for d²y/dx² (i.e. y''), so we can do implicit differentiation again.

But first, let me rewrite this in terms of y', to keep the notation consistent with the question:

16x¹ + 2y * y' = 0

Before we do this, I want you to think about what we have in the above equation.

We have 16x, which is easy to deal with.

But then we have 2y * y'.

What is y? Some function of x, but we don't know what it is.

What is y'? it is another function of x, but we also don't know what it is.

When taking the derivative of two functions multiplied by each other we have a rule we can use.

The product rule.

(u * v)' = u' * v + v' * u.

Before we used the chain rule, but now we have to use the product rule.

Let's give it a go!

(16x)' + (2y * y')' = 0

16 + (2y)'*y' + 2y*(y')' = 0

16 + (2 * y')*y' + 2y * y'' = 0.

It is a little hard to see what is going on with all of those teeny prime marks, but let's go through it step by step.

(16x)' is just 16, since we are taking the derivative with respect to x.

(2y)' can be thought of in two different ways.

First, we can think of it in terms of the chain rule.

The derivative of 2y is just 2, and then we multiply by the derivative of y with respect to x, which is dy/dx, or y' as we denote it above.

So (2y)' = 2*1*y'

Second, we could just think of (y)' = y', so (2y)' = 2 * y'.

And (y')' is just y'', as expected. (the derivative of the derivative is the second derivative by definition)

So we have:

16 + (2 * y')*y' + 2y * y'' = 0.

Simplifying yeilds:

16 + 2 * (y')² + 2y * y'' = 0

Then solving for y'':

2y * y'' = -16 - 2* (y')²

Dividing by 2y:

y'' = (-16 - 2* (y')²)/(2y)

note that you could / probably should cancel the 2 out:

y'' = (-8 - (y')²)/(y)

So this should be the answer.

We could solve and integrate the earlier equation of 16x¹ + 2y * y' = 0 for y, and then take the derivative twice to verify that this answer is correct, but that would take a lot of time.

I hope this helps you! Please let me know if you have any questions, or if something is unclear.

I do not believe I made any mistakes, but if you notice any feel free to mention it in a reply below.