a. Answer: [tex]\bold{-\dfrac{5}{9}}[/tex]
Step-by-step explanation:
[tex]x^2+2y^3=\dfrac{3}{xy}\\\\\text{Cross multiply:}\quad x^3y+2xy^4=3\\\\\text{Differentiate each term separately using ab=a'b+ab':}\\x^3y\qquad \qquad \qquad \qquad \qquad 2xy^4\\a=x^3\quad a'=3x^2\qquad \qquad a=2x\quad a'=2\\b=y\ \quad b'=y'\qquad \qquad \quad b=y^4\quad b'=4y^3y'\\\\x^3y'+3x^2y+8xy^3y'+2y^4=0\\\underline{-x^3y'\quad \quad -8xy^3y'\qquad \quad -x^3y'-8xy^3y'}\\.\qquad \quad 3x^2y\qquad \qquad +2y^4=-x^3y'-8xy^3y'[/tex]
[tex]\text{Factor out y' on the right side:}\\3x^2y+2y^4=-y'(x^3+8xy^3)\\\\\text{Divide both sides by}\ -(x^3+8xy^3):\\-\dfrac{3x^2y+2y^4}{x^3+8xy^3}=y'\\\\\text{Substitute x = 1, y = 1 to find the derivative at that coordinate:}\\-\dfrac{3x(1)^2(1)+2(1)^4}{(1)^3+8(1)(1)^3}=y'\\\\\boxed{-\dfrac{5}{9}=y'}[/tex]
b. Answer: -1
Step-by-step explanation:
[tex]y=\dfrac{x+y}{x-y}\\\\\text{Cross Multiply:}\\xy-y^2=x+y\\\\\text{Differentiate using ab=a'b+ab'}:\\a=x\qquad a'=1\\b=y\qquad b'=y'\\\\.\quad xy'+y-2y\cdot y'=1+y'\\\underline{-xy'\ -1+2y\cdot y'\quad -1-xy'+2y\cdot y'}\\.\qquad y-1\qquad \qquad =y'-xy'+2y\cdot y'\\\\\text{Factor out y' from the right side:}\\y-1=y'(1-x+2y)[/tex]
[tex]\text{Divide 1-x+2y from both sides:}\\\dfrac{y-1}{1-x+2y}=y'\\\\\text{Substitute x = 6 and y = 2 to find the derivative at that coordinate:}\\\dfrac{(2)-1}{1-(6)+2(2)}=y'\\\\\dfrac{1}{-1}=y'\\\\\boxed{-1=y'}[/tex]
