How to find implicit differentiation to find dy/dx in ti84

[carlycakes]

i found a program for an ti 83/84 where you can use to find the implicit differentiation.
its called the 'The Ultimate Calculus Collection!!!' and i downloaded from ticalc.com --> http://www.ticalc.org/pub/83plus/basic/ ... uscalc.zip
i need help typing in a the function into the program to check my answer!
its asking for 'Equality' and i have no idea what that means....and my equation is...
1) tan(x+y) = x and
2) x^3 + 3x^2 + y^3 = 8

 

[galactus]

Re: How to find implicit differentiation to find dy/dx in ti

I am not familiar with that particular program, but I would assume that the equality means you have to enter something in for the right side of the equals sign.

But, to do it the old-fashioned way:

\(\displaystyle tan(x+y)=x\)

\(\displaystyle sec^{2}(x+y)(1+y')=1\)

\(\displaystyle sec^{2}(x+y)+y'sec^{2}(x+y)=1\)

\(\displaystyle y'sec^{2}(x+y)=1-sec^{2}(x+y)\)

\(\displaystyle y'=\frac{1-sec^{2}(x+y)}{sec^{2}(x+y)}\)

\(\displaystyle y'=\frac{1}{sec^{2}(x+y)}-1\)

\(\displaystyle y'=cos^{2}(x+y)-1\)

\(\displaystyle y'=-sin^{2}(x+y)\)

There is a step by step for that one to help with others. Is that what you got?.

With my Voyage 200, I have to enter in \(\displaystyle ImpDif(tan(x+y)=x,x,y)\)

Your program may be similar.

 

[mmm4444bot]

:idea: Did you look at the README.TXT file ?

The developer's name is Chip. He provides his e-mail address; please contact Chip for technical support.