Trouble understanding with taking the second derivative

Question

Problem Statement

In the following image, I cannot understand why taking the derivative of $\sqrt{1-y^2}$ that you must do $\frac{\frac{d}{dx}[1-y^2]}{2\sqrt{1-y^2}}$ in order to get to the solution.

Here are the steps (partially)

The step that I do not understand is the one circled in red.

Conclusion

As you saw, the step circled in red is the part I do not understand. To me, it makes no sense how you get to that step. It would be nice if someone could explain how to get to that step from the previous one.

Is $(\sqrt{u})'=\frac{u'}{2\sqrt u}$ the formula you are looking for? — Balloon, Jan 27 '20 at 15:32
If $u$ is a positive differentiable function, then the function $\sqrt{u}$ is differentiable with derivative $(\sqrt{u})'=\frac{u'}{2\sqrt{u}}$. In your case, i.e. with $u=1-y^2$ and the notation $u'=\frac{d}{dx}u$, you get what is written. — Balloon, Jan 27 '20 at 15:36
Ok, great! It makes sense now. Could you please post your answer? — Mathaholic, Jan 27 '20 at 15:37
If you are okay with the one of Andrew, there is no problem for me! — Balloon, Jan 27 '20 at 15:39

score 2 · Accepted Answer · answered Jan 27 '20 at 15:37

2

We know $$\frac{d}{dx}\sqrt{x}=\frac1{2\sqrt{x}}.$$ By the chain rule, we have $$\frac{d}{dx}\sqrt{u}=\frac1{2\sqrt{u}}\cdot\frac{du}{dx}=\frac{\frac{du}{dx}}{2\sqrt{u}}.$$

In your example, we have $u=1-y^2$.

answered Jan 27 '20 at 15:37

Andrew Chin

7,389

score 0 · Answer 2 · answered Feb 03 '20 at 09:30

The step you don't understand is hidden in some of the details associated with 2 applications of the Chain Rule.

$\frac {d}{dx} \left( \frac {dy}{dx} \right) = \frac {d}{dx} (1-y^2)^{1/2}$ (Eq01)

$\frac {d^2y}{dx^2} = \frac {d}{dx} (1-y^2)^{1/2}$

Now, let $\mu=(1-y^2)$ (Eq02)

We now have $\frac {d^2y}{dx^2} = \frac {d}{dx} \mu^{1/2}$ (Eq03)

We now have a mis-match of the Derivative Operator (x), and the function ($\mu$). The Chain Rule essentially says you can change the x to $\mu$ ONLY IF you multiply by d$\mu$/dx. So Eq03 becomes

$\frac {d^2y}{dx^2} = \left(\frac {d}{d\mu} \mu^{1/2}\right)\cdot\left(\frac {d\mu}{dx}\right)$ (Eq04)

The first parenthesis on the RHS of Eq04 is straightforward derivative of a monomial

$\left(\frac {d}{d\mu} \mu^{1/2}\right)=\frac {1}{2}\mu^{\frac {1}{2}-1}=\frac {1}{2\mu^{1/2}}$ (Eq04a)

The second parenthesis on the RHS of Eq04 is determined by applying the Derivative Operator to Eq02, the variable substitution.

$\frac {d}{dx}\mu=\frac {d}{dx}(1-y^2)\Rightarrow \frac {d\mu}{dx}=\frac {d}{dx}(1)-\frac {d}{dx}y^2\Rightarrow\frac {d\mu}{dx}=0-\frac {d}{dx}y^2$

Now, here again, we have a mis-match and must use the Chain Rule again.

$\frac {d\mu}{dx}=-\frac {d}{dx}y^2=-\frac {d}{dy}y^2\frac {dy}{dx}=-2y\frac {dy}{dx}$ Therefore,

$\frac {d\mu}{dx}=-2y\frac {dy}{dx}$ (Eq04b)

Substituting (Eq04a) and (Eq04b) into (Eq04):

$\frac {d^2y}{dx^2} = \left(\frac {1}{2\mu^{1/2}}\right)\cdot\left(-2y\frac {dy}{dx}\right)=\frac {-2y\frac {dy}{dx}}{2\mu^{1/2}}=\frac {-2y\frac {dy}{dx}}{2(1-y^2)^{1/2}}$

$\frac {d^2y}{dx^2} = \frac {-y\frac {dy}{dx}}{(1-y^2)^{1/2}}$

Trouble understanding with taking the second derivative

2 Answers2