Expert: Abe Mantell - 1/14/2005

Question
Thanks
got one more quick one no one seems to know
using implicipt differentiation find the equation of the line tangent to the curve x^3+2xy+y^3=13 at point (1,2)
Thanks Again
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Followup To
Question -
I totally dont get it
Find the function f(x)=x^k and a function g such as that f(g(x))=h(k)=(3x+x^2)^(1/2)
huh

Answer -
Hello Georgia,

There are many possible answers...but the most
"obvious" one is:
k=1/2 and g(x)=3x+x^2...so the composition of the two
gives the desired function...see?

TTYL, Abe

Answer
Hello again...

Differentiate with respect to x, so we get:
3x^2+2xy'+2y+3y^2y'=0
(using the product rule for 2xy and the chain-rule)
now solve for y'...
y'(2x+3y^2)=-2y-3x^2
so y'=(-2y-3x^2)/(2x+3y^2)

So, at (1,2) y'=(-4-3)/(2+14)=-7/16

Thus, the tangent line is given by:
y-2=(-7/16)(x-1)...I think you can finish it off now...yes?

TTYL, Abe