$\frac { d y } { d x } \text { at the point } ( 0,20 ) \text { for the equation } x = 14 \ln \left( y ^ { 2 } - 399 \right) \text {. }$

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The Answer of $\frac { d y } { d x } \text { at the point } ( 0,20 ) \text { for the equation } x = 14 \ln \left( y ^ { 2 } - 399 \right) \text {. }$

dydx at the point (0,20) for the equation x=14ln⁡(y2−399). \frac { d y } { d x } \text { at the point } ( 0,20 ) \text { for the equation } x = 14 \ln \left( y ^ { 2 } - 399 \right) \text {. }dxdy​ at the point (0,20) for the equation x=14ln(y2−399).