Find dy/dx by implicit differentiation

Question

xy³

^________ =1+y⁴

sec(y)

Answer 1

It's a bit complicated but here goes:

Let's first work with the left side of the equation:

Let u = xy³ then u' = (x)'y³ + x(y³)' or u' = y³ + x(3y²)(dy/dx) [using the product rule]

Let v = sec(y) then v' = sec(y)tan(y)(dy/dx)

Using the quotient rule (u/v)' = (u'v - uv')/v²

(u/v)' ={[y³ + x(3y²)(dy/dx)]sec(y) - (xy³)(sec(y)tan(y)(dy/dx))}/sec²(y)

(u/v)' ={[y³ + x(3y²)(dy/dx)] - (xy³)(tan(y)(dy/dx))}/sec(y)

Now, working with the right side of the equation:

(1 + y⁴)' = 0 + 4y³(dy/dx)

(1 + y⁴)' = 4y³(dy/dx)

So putting them together we get:

{[y³ + x(3y²)(dy/dx)] - (xy³)(tan(y)(dy/dx))}/sec(y) = 4y³(dy/dx)

Multiplying both sides by sec(y) we get:

y³ + x(3y²)(dy/dx) - (xy³)(tan(y)(dy/dx)) = 4y³(dy/dx)sec(y)

y³ + 3xy²(dy/dx) - (xy³)tan(y)(dy/dx) = 4y³(dy/dx)sec(y)

Moving all the "dy/dx" terms to the left side of the equation and the other term to the right we get:

3xy²(dy/dx) - (xy³)tan(y)(dy/dx) - 4y³(dy/dx)sec(y) = -y³

Factoring out dy/dx we get:

(dy/dx)[3xy² - xy³tan(y) - 4y³sec(y)] = -y³

Dividing both sides by the "[3xy² - xy³tan(y) - 4y³sec(y)]" we get:

dy/dx = -y³/[3xy² - xy³tan(y) - 4y³sec(y)]

If you want to get rid of the negative in front of the y³ you could multiply top and bottom by -1 to get:

dy/dx = y³/(xy³tan(y) + 4y³sec(y) - 3xy²)

Cancelling y² from the terms on the bottom gives:

dy/dx = y/(xytan(y) + 4ysec(y) - 3x)