28a) Answer: [tex]\bold{y'=-\dfrac{5}{3}}[/tex]
Step-by-step explanation:
[tex]5x + 3y = 12\\\\\\\text{take the derivative of both sides with respect to x:}\\(5x + 3y)\dfrac{dy}{dx}=(12)\dfrac{dy}{dx}\\\\\rightarrow \quad 5+3y'=0\\\\\\\text{Subtract 5 from both sides and divide 3 from both sides to solve for y':}\\\rightarrow \quad 3y'=-5\\\\\rightarrow \quad y'=-\dfrac{5}{3}[/tex]
28b) Answer: [tex]\bold{y'=-\dfrac{1-10(2x+3y)^4}{15(2x+3y)^4}}[/tex]
Step-by-step explanation:
[tex](2x+3y)^5=x+1\\\\\\\text{Take the derivative of both sides with respect to x:}\\(2x+3y)^5\dfrac{dy}{dx}=(x+1)\dfrac{dy}{dx}\\\\\rightarrow \quad 5(2x+3y)^4\cdot (2+3y')=1+0\\\\\\\text{Divide both sides by }5(2x+3y)^4}:\\2+3y'=\dfrac{1}{5(2x+3y)^4}\\\\\\\text{Subtract 2 from both sides and divide both sides by 3 to solve for y'}:\\3y'=\dfrac{1}{5(2x+3y)^4}-2\\\\y'=\dfrac{1}{15(2x+3y)^4}-\dfrac{2}{3}[/tex]
[tex]\text{Give them a common denominator}:\\y'=\dfrac{1}{15(2x+3y)^4}-\dfrac{2}{3}\bigg(\dfrac{5(2x+3y)^4}{5(2x+3y)^4}\bigg)\\\\\\y'=\dfrac{1-10(2x+3y)^4}{15(2x+3y)^4}[/tex]
