Find dy/dx at the given point for the equation.
x = 9 ln(y^2 − 48), (0, 7)
Thanks for the help, really appreciated! :)
x = 9 ln(y^2 − 48), (0, 7)
Thanks for the help, really appreciated! :)
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Okay, I'll try make this as clear as possible:
Step1: 1 = 9( 2y dy/dx / y^2 - 48 )
Step2: 1 = 18ydy/dx /y^2 - 48
(Rearrange to get in terms of dy/dx)
Step3: y^2 - 48 = 18y dy/dx
Step 4: dy/dx = (y^2 - 48) /18y
Now plug in the value for y in the equation which we have from the given point : 7
dy/dx = (7^2-48) / 18(7)
dy/dx = 1 /126
Step1: 1 = 9( 2y dy/dx / y^2 - 48 )
Step2: 1 = 18ydy/dx /y^2 - 48
(Rearrange to get in terms of dy/dx)
Step3: y^2 - 48 = 18y dy/dx
Step 4: dy/dx = (y^2 - 48) /18y
Now plug in the value for y in the equation which we have from the given point : 7
dy/dx = (7^2-48) / 18(7)
dy/dx = 1 /126