Implicit Double Derivatives: x^2 + y ^ 2 = 25

[grapz]

1) Find the second derivative of x^2 + y ^ 2 = 25

I can only find the first derivative i can't find the second.

2) Find the second derivative of y = x^2 y^3 + xy

I actually have no clue how to find the second derviative. This sort of question is going to be on a test, but my teacher didn't cover it. So please explain it step by step. Thank you!

 

[stapel]

The second derivative is the derivative of the first derivative. So apply the exact same methods to the first derivative as you did to the original equation.

If you get stuck, please reply showing your work and reasoning. Thank you.

Eliz.

Edit: Ne'mind: complate answer posted below.

 

[soroban]

Re: Implicit Double Derivatives

Hello, grapz!

You really have no clue about how to find a second derivative?

\(\displaystyle x^2 + y ^ 2\:=\:25\)
Find the second derivative.

Differentaite implicitly: \(\displaystyle \:2x\,+\,2y\left(\frac{dy}{dx}\right)\:=\:0\)

Solve: \(\displaystyle \L\:2y\left(\frac{dy}{dx}\right)\:=\:-2x\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:-\frac{x}{y}\;\;\) Did you get this far?


Differentiate again: \(\displaystyle \L\:\frac{d^2y}{dx^2} \;=\;-\frac{y\cdot1\,-\,x\left(\frac{dy}{dx}\right)}{y^2}\)


Substitute the first derivative: \(\displaystyle \L\:\frac{d^2y}{dx^2} \;=\;-\frac{y \,-\,x\left(-\frac{x}{y}\right)}{y^2} \;=\;-\frac{y\,+\,\frac{x^2}{y}}{y^2}\)

Multiply top and bottom by \(\displaystyle y:\;\;\L\frac{d^2y}{dx^2}\;=\;-\frac{y^2\,+\,x^2}{y^3}\)


But the original equation says: \(\displaystyle x^2\,+\,y^2\:=\:25\)
. . We can replace the numerator with \(\displaystyle 25\). **

Therefore: \(\displaystyle \L\:\fbox{\frac{d^2y}{dx^2} \;= \;-\frac{25}{y^3}}\)


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**

This type of substitution turns up quite often
. . with implicit second derivatives.
Watch for it!

 

[grapz]

i see, i got to it, just didnt' subsite the 25 in, and was wondering why.