Implicit Differentials

[Crystal24]

Hi I"m in brief calculus and I need to help solving
x²/16 + y² /25 = 1
y(1)=4.84
y'(1) = ?

Do I go about it this way?
(16(x²)' - (x²)(16)')/16² + (25)(y²)' - y²(25)')/25² = (1)'
then its 16(2x)-(x²)(0)/256 + (25)(2y)(y') - y²(0)/625 = 0
simplify: 32x/256 + 50y(y')/625 = 0
not sure where to go from there...... Help please

 

[pka]

Hi I"m in brief calculus and I need to help solving
x²/16 + y² /25 = 1
y(1)=4.84
y'(1) = ?

\(\displaystyle \dfrac{x}{8}+\dfrac{2yy^{\prime}}{25}=0\)

\(\displaystyle 50x+32yy^{\prime}=0\)

\(\displaystyle y^{\prime}=\dfrac{-25x}{16y}\)