Answer:
a. dy/dx = -2/3
b. dy/dx = -28
Step-by-step explanation:
One way to do this is to assume that x and y are functions of something else, say "t", then differentiate with respect to that. If we write dx/dt = x' and dy/dt = y', then the required derivative is y'/x' = dy/dx.
a. x'·y^3 +x·(3y^2·y') = 0
y'/x' = -y^3/(3xy^2) = -y/(3x)
For the given point, this is ...
dy/dx = -2/3
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b. 2x·x' +x^2·y' -2x'·y^3 -2x·(3y^2·y') + 0 = 2x' + 2y'
y'(x^2 -6xy^2 -2) = x'(2 -2x +2y^3)
y'/x' = 2(1 -x +y^3)/(x^2 +6xy^2 -2)
For the given point, this is ...
dy/dx = 2(1 -0 +27)/(0 +0 -2)
dy/dx = -28
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The attached graphs show these to be plausible values for the derivatives at the given points.
Answer:
12-2=10 b=10
Step-by-step explanation:
Part a I am pretty sure it is a pattern so 4 plus 2 is 6, 6 plus 3 is nine and then 9 plus 4 is 13 Then 13 plus 5 is 18
Answer is 18
In part b he will not finish his work he has to get to 274 and he only gets to about 50 something
Hope this helps