Whats the square root of 9
1 answer:
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When x=-1: 
Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?
The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.
From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4
and solve for x'.
Hope that helps!
Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?
The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.
From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4
and solve for x'.
Hope that helps!
