Question

Finding the 2nd derivative using Implicit differentiation

Refer to this if you want to find out about the first derivative: https://questioncove.com/study#/updates/5fb6d473283658df98d49000

Answer 1

Now lets take the example of
\[x^2-xy+y^2=1\]
we are going to find \[d^2y/dx^2\]
To do this we need to first find the first derivative. I will not go over this fully because that was what my other tutorial was for, refer to this link here
https://questioncove.com/study#/updates/5fb6d473283658df98d49000
So the first derivative we find to be
\[(2x-y)/(x-2y)\]
This is because we use the rules and get the derivative of
\[2x - x \times dy/dx - y + 2y \times dy/dx\]
And then solving for \[dy/dx \] we get, \[dy/dx = (2x-y)/(x-2y)\]
Now we want to find the second derivative using implicit differentiation. To do this we want to find the derivative of this derivative function above. Now we use the Quotient rule to find the derivative.
\[((x-2y)(2-dy/dx) - ((2x-y)(1-2 \times dy/dx)))/(x-2y)^2\]
What we did above is apply Quotient rule, which is for h = f/g
the derivative is \[(g \times f' - f \times g')/g^2\]
So now we just have to simplify.
distribute:
\[(2x-4y-x \times dy/dx + 2y \times dy/dx - 2x +y +4x \times dy/dx - 2y \times dy/dx)/(x-2y)^2\]
We can cancel some terms
\[(-4y-x \times dy/dx + y + 4x \times dy/dx)/(x-2y)^2\]
This can further simplify down to
\[(-3y + 3x \times dy/dx)/(x-2y)^2\]
Well we are not done yet, what do we plug in for dy/dx?
Well we know that \[dy/dx = (2x-y)/(x-2y)\]
So substituting that in for dy/dx we now have
\[(-3y + 3x \times (2x-y)/(x-2y))/(x-2y)^2\]
Now we just have to simplify this. We have
\[(-3y + ((6x^2-3xy)/(x-2y)))/(x-2y)^2\]
Move the x-2y and multiply with (x-2y)^2 to get (x-2y)^3 and -3y = 6y^2-3xy
Simplifying again, we have \[d^2y/dx^2= (6x^2-6xy+6y^2)/(x-2y)^3\]
This is our final answer, more examples to come.

Answer 2

@hero @umm @shadow please move to correct subject? it won't let me switch to tutorial