Question

Hi, Can I get some help with this calculus homework? Please show your work! Thanks! #28 a and b ​

Answer 1

28a) Answer:
`\bold{y'=-(5)/(3)}`

Explanation:

`5x + 3y = 12\\\\\\\text{take the derivative of both sides with respect to x:}\\(5x + 3y)(dy)/(dx)=(12)(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'=-(5)/(3)`

28b) Answer:
`\bold{y'=-(1-10(2x+3y)^4)/(15(2x+3y)^4)}`

Explanation:

`(2x+3y)^5=x+1\\\\\\\text{Take the derivative of both sides with respect to x:}\\(2x+3y)^5(dy)/(dx)=(x+1)(dy)/(dx)\\\\\rightarrow \quad 5(2x+3y)^4\cdot (2+3y')=1+0\\\\\\\text{Divide both sides by }5(2x+3y)^4}:\\2+3y'=(1)/(5(2x+3y)^4)\\\\\\\text{Subtract 2 from both sides and divide both sides by 3 to solve for y'}:\\3y'=(1)/(5(2x+3y)^4)-2\\\\y'=(1)/(15(2x+3y)^4)-(2)/(3)`

`\text{Give them a common denominator}:\\y'=(1)/(15(2x+3y)^4)-(2)/(3)\bigg((5(2x+3y)^4)/(5(2x+3y)^4)\bigg)\\\\\\y'=(1-10(2x+3y)^4)/(15(2x+3y)^4)`