Respuesta :
Answer:
See below
Step-by-step explanation:
Problem A
[tex]\displaystyle 5x+3y=12\\\\\frac{d}{dx}\bigr(5x+3y)= \frac{d}{dx}\bigr(12)\\\\5+3\frac{dy}{dx}=0\\ \\3\frac{dy}{dx}=-5\\ \\\frac{dy}{dx}=-\frac{5}{3}[/tex]
Problem B
[tex]\displaystyle (2x+3y)^5=x+1\\\\\frac{d}{dx}\bigr(2x+3y)^5=\frac{d}{dx}\bigr(x+1)\\ \\\biggr(2+3\frac{dy}{dx}\biggr)\biggr(5(2x+3y)^4\biggr)=1 \leftarrow \text{Chain Rule}\\\\\biggr(10+15\frac{dy}{dx}\biggr)(2x+3y)^4=1\\\\10+15\frac{dy}{dx}=\frac{1}{(2x+3y)^4}\\ \\15\frac{dy}{dx}=\frac{1}{(2x+3y)^4}-10\\ \\15\frac{dy}{dx}=\frac{1}{(2x+3y)^4}-\frac{10(2x+3y)^4}{(2x+3y)^4}\\\\15\frac{dy}{dx}=\frac{1-10(2x+3y)^4}{(2x+3y)^4}\\ \\\frac{dy}{dx}=\frac{1-10(2x+3y)^4}{15(2x+3y)^4}[/tex]
Remember that when doing implicit differentiation, you treat y as a constant and write dy/dx. Also, remember that chain rule is the inside derivative times the outside derivative!
