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!
Hello :
cos60 = cos(-60)=0.5
Answer:
222
Step-by-step explanation:
1114-5646
Answer:
x = 1 or x = −6
Step-by-step explanation:
Step 1: Subtract 18 from both sides.
3x2+15x−18=18−18
3x2+15x−18=0
Step 2: Factor left side of equation.
3(x−1)(x+6)=0
Step 3: Set factors equal to 0.
x−1=0 or x+6=0
x=1 or x=−6
A function is differentiable if you can find the derivative at every point in its domain. In the case of f(x) = |x+2|, the function wouldn't be considered differentiable unless you specified a certain sub-interval such as (5,9) that doesn't include x = -2. Without clarifying the interval, the entire function overall is not differentiable even if there's only one point at issue here (because again we look at the entire domain). Though to be fair, you could easily say "the function f(x) = |x+2| is differentiable everywhere but x = -2" and would be correct. So it just depends on your wording really.