equation of the tangent line

Jan 2010
79
1
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.



I've found the derivative of the function to be:
\(\displaystyle (x (-4 x^2-4 y^2+25))/(y (4 x^2+4 y^2+25))\)

To find the tangent line at (3,1), you just plug in the values to find the slope, correct? I've done this and can't seem to find the right tangent line at that point..
 
Jan 2010
564
242
Kuwait
Assuming your derivative is correct, so we have:

\(\displaystyle y'=\frac{x(-4x^2-4y^2+25)}{y(4x^2+4y^2+25}\)

Find \(\displaystyle y'(3,1)\) to get the slope ..

then substitute this in the well-known equation of any line ..
 
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