Question

use implicit differentiation to find an equation of the tangent of the curve at the given point. ( will post the problem in a second)

Answer 1

\[2(x ^{2}+y ^{2})^{2}=25(X ^{2}+y ^{2}) \]

the point is (3,1)

Answer 2

\[\frac{ d }{ dx }[2(x^{2} + y^{2})^2]=25 \frac{ d }{ dx }[(x^2+y^2]\]

The chain rule should be used in the left side of the equation

\[4(x^2+y^2)(x^2+y^2)'\]=\frac{ d }{ dx }[25(x^2+y^2]\]


for the left side of the equation, we could use the power rule for x^2 and the chanin rule for y^2
4(2x+2y(y)'=25(2x+2y(y)')

Answer 3

\[4(x^2+y^2)(2x+2y(y)')=25(2x+2y(y)')\]

Answer 4

Now, solve for y prime (y)' :)

Answer 5

Did i make it clear? :)

Answer 6

i think so. I am trying to solve for y prime real fast

Answer 7

Ok ^.^

Answer 8

I have no idea. I keep trying but its not coming out right. can you divide both sides by 2x+2y(yprime)?

Answer 9

which would leave you with no yprimes? allowing you to solve for y and get your tangent line?

Answer 10

I'll just work the problem through to see if you got it right
\[4(x^2+y^2)(2x+2y(z))=25(2x+2yz)\]
Think of y prime as z, and you are trying to solve for it

\[=(8x^2+4y^2)(2x+2yz)=50x+50yz\]
expand (8x^2+4y^2)(2x+2yz)
\[=8x^3+8xy^2+8x^2yz+8y^3z=50x+50yz\]
subtract 50yz from both sides
\[=-50y+8x^2y^2-50yz+8x^2yz+8y^3z=50x\]
factor z out
\[=(-50y+8x^2y+8y^3)z=50x-8x^3-8xy^2\]
now, divide everything, except z to the left side of the equation to isolate z by itself
\[z=\frac{ (50x-8x^3-8xy^2) }{ -50y+8x^2y+8y^3 }\]

Answer 11

i wasnt factoring.. that might have been my first issue haha

Answer 12

Yeah, haha factoring is really important when differentiating
algebra is the main tool in calculus, if something was wrong, blame it on the algebra part :)

Answer 13

alright I think i got it! and thanks!

Answer 14

No problem, and good luck!