if x^2 + y^2 + 6x - 16 = 0
find dy/dx when x = 0
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is that correct differentiate with respect to x
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Given : x^2 + y^2 + 6x - 16 = 0
To Find : dy/dx when x = 0
Solution:
x² + y² + 6x - 16 = 0
=> 2x + 2y. dy/dx + 6 = 0
=> x + y. dy/dx + 3 = 0
=> dy/dx = - ( x + 3)/y
x² + y² + 6x - 16 = 0
x = 0
=> 0 + y² + 0 - 16 = 0
=> y = ± 4
dy/dx = - ( x + 3)/y
= - (0 + 3) / ± 4
= ± 3/4
dy/dx when x = 0 = ± 3/4
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