$\frac { d y } { d x } \text { at the point } ( 0,20 ) \text { for the equation } x = 14 \ln \left( y ^ { 2 } - 399 \right) \text {. }$
A) $\frac { 1 } { 560 }$
B) $- \frac { 57 } { 80 }$
C) $- \frac { 1 } { 560 }$
D) $\frac { 57 } { 80 }$
E) $\frac { 1 } { 280 }$

Question

$\frac { d y } { d x } \text { at the point } ( 0,20 ) \text { for the equation } x = 14 \ln \left( y ^ { 2 } - 399 \right) \text {. }$ 
A) $\frac { 1 } { 560 }$
B) $- \frac { 57 } { 80 }$
C) $- \frac { 1 } { 560 }$
D) $\frac { 57 } { 80 }$
E) $\frac { 1 } { 280 }$

Quizplus · Accepted Answer

Explanation:To find the derivative of the given equation, we can use implicit differentiation. Differentiating both sides of the equation with respect to x, we get:$\frac { d } { d x } [ x ] = \frac { d } { d x } [ 14 \ln ( y ^ { 2 } - 399 ) ]$$\Rightarrow 1 = \frac { 14 } { y ^ { 2 } - 399 } \cdot \frac { d } { d x } [ y ^ { 2 } - 399 ]$$\Rightarrow 1 = \frac { 14 } { y ^ { 2 } - 399 } \cdot 2y \frac { d y } { d x }$$\Rightarrow \frac { d y } { d x } = \frac { 1 } { 14y } \cdot \frac { y ^ { 2 } - 399 } { 2y }$$\Rightarrow \frac { d y } { d x } = \frac { y - 28 } { 28y }$At the point (0, 20), we have:$\frac { d y } { d x } = \frac { 20 - 28 } { 28(20) } = \frac { -8 } { 560 } = - \frac { 1 } { 560 }$Therefore, the correct answer is A.

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).

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