Question

Find dy/dx by implicit differentiation.

a. 5x + 3y = 12

b. (2x + 3y)^ 5 = x + 1


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Answer 1

See below

Explanation:

Problem A

`\displaystyle 5x+3y=12\\\\(d)/(dx)\bigr(5x+3y)= (d)/(dx)\bigr(12)\\\\5+3(dy)/(dx)=0\\ \\3(dy)/(dx)=-5\\ \\(dy)/(dx)=-(5)/(3)`

Problem B

`\displaystyle (2x+3y)^5=x+1\\\\(d)/(dx)\bigr(2x+3y)^5=(d)/(dx)\bigr(x+1)\\ \\\biggr(2+3(dy)/(dx)\biggr)\biggr(5(2x+3y)^4\biggr)=1 \leftarrow \text{Chain Rule}\\\\\biggr(10+15(dy)/(dx)\biggr)(2x+3y)^4=1\\\\10+15(dy)/(dx)=(1)/((2x+3y)^4)\\ \\15(dy)/(dx)=(1)/((2x+3y)^4)-10\\ \\15(dy)/(dx)=(1)/((2x+3y)^4)-(10(2x+3y)^4)/((2x+3y)^4)\\\\15(dy)/(dx)=(1-10(2x+3y)^4)/((2x+3y)^4)\\ \\(dy)/(dx)=(1-10(2x+3y)^4)/(15(2x+3y)^4)`

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!