Question

5. (3.5pts) Given the equation y = y²x+2 cos(x) find . Show each step of your work.
dx

Answer 1

`(dy)/(dx)=(y^2-2 \sin(x))/(1-2xy)`

Explanation:

Given equation:

`y=y^2x+2 \cos(x)`

To find the derivative of the given equation, use implicit differentiation.

Add d/dx in front of each term:

`(d)/(dx)\:y =(d)/(dx)\:y^2x+(d)/(dx)\:2 \cos(x)`

Differentiate terms in x only (and constant terms) with respect to x:

`(d)/(dx)\:y =(d)/(dx)\:y^2x-2 \sin(x)`

Use the chain rule to differentiate terms in y only:

(in practice this means to differentiate with respect to y then place dy/dx on the end)

`(dy)/(dx)\: =(d)/(dx)\:y^2x-2 \sin(x)`

Use the product rule on the term in x and y:

`\textsf{let }u=y^2 \implies (du)/(dx)=2y (dy)/(dx)`

`\textsf{let }v=x \implies (dv)/(dx)=1`

`\implies u(dv)/(dx)+v(du)/(dx)=y^2+2xy (dy)/(dx)`

Therefore:

`(dy)/(dx) =y^2+2xy (dy)/(dx)-2 \sin(x)`

Rearrange to make dy/dx the subject:

`(dy)/(dx)-2xy (dy)/(dx) =y^2-2 \sin(x)`

`(dy)/(dx)(1-2xy)=y^2-2 \sin(x)`

`(dy)/(dx)=(y^2-2 \sin(x))/(1-2xy)`