Question

Hello, Can I get some help with #30 a and b, please? Don't forget to show your work.

Answer 1

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 2

a. Answer:  $\bold{-\dfrac{2}{3}}$

Step-by-step explanation:

$xy^3=8\\\\\text{Differentiate using ab=a'b+ab'}:\\a=x\qquad a'=1\\b=y^3\qquad b'=3y^2\cdot y'\\\\y^3+3xy^2\cdot y'=0\\\\\text{Subtract }3xy^2\cdot y'\ \text{from both sides}:\\y^3=-3xy^2\cdot y'\\\\\text{Divide }-3xy^2\ \text{from both sides}:\\-\dfrac{y^3}{3xy^2}=y'\\\\\text{Substitute x = 1 and y = 2 to find the derivative at that coordinate:}\\-\dfrac{(2)^3}{3(1)(2)^2}=y'\\\\\\-\dfrac{8}{12}=y'\\\\\\\boxed{-\dfrac{2}{3}=y'}$

b. Answer:  -28

Step-by-step explanation:

$x^2y-2xy^3+6=2x+2y\\\\\text{Differentiate each term separately using ab=a'b+a'b}:\\.\quad \qquad x^2y\qquad \qquad \qquad \qquad \qquad -2xy^3\\a=x^2\qquad a'=2x\qquad \qquad a=-2x\qquad a'=-2\\b=y\qquad \ b'=y'\qquad \qquad \ b=y^3\qquad \ b'=3y^2\cdot y'\\\\2xy+x^2\cdot y'-2y^3-6xy^2\cdot y'+0=2+2y'$

$\text{Subtract }x^2y\cdot y',\ \text{Subtract 2, and Add }6xy^2\cdot y'\ \text{on both sides:}\\2xy-2y^3-2=2y'-x^2y'+6xy^2\cdot y'\\\\\text{Factor out y' from the right side:}\\2xy-2y^3-2=y'(2-x^2+6xy^2)\\\\\text{Divide both sides by }2-x^2+6xy^2:\\\dfrac{2xy-2y^3-2}{2-x^2+6xy^2}=y'\\\\\text{Substitute x = 0 and y = 3 to find the derivative at that coordinate:}\\\dfrac{2(0)(3)-2(3)^3-2}{2-(0)^2+6(0)(3)^2}=y'\\\\\\\dfrac{-56}{2}=y'\\\\\boxed{-28=y'}$