Question

How do you find the derivative of `x+tan(xy)=0`?

Answer 1

`dy/dx={(1+x^2)arc tanx-x}/{x^2(1+x^2)}`

Explanation:

`x+tan(xy)=0 rArr tan(xy)=-x rArr xy=arc tan(-x)`

`rArr xy=-arc tanx`

`y=-(arc tanx)/x`

By the Quotient Rule , then,

`dy/dx=-[{xd/dx(arc tanx)-(arc tanx)d/dx(x)}/x^2]`

`=-[{x/(1+x^2)-arc tanx}/x^2]`

`:. dy/dx={(1+x^2)arc tanx-x}/{x^2(1+x^2)}`

Answer 2

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

Explanation:

Use implicit differentiation and algebra to get `dy/dx` on one side of the equation

1. Differente both sides of the equation
   `color(red)(d/dx)(x)+color(red)(d/dx)(tan(xy))=color(red)(d/dx)(0)`

2. `d/dx(x)=1` and `d/dx(0)=0`
   `color(red)(1)+d/dx(tan(xy))=color(red)(0)`

3. The derivative of the `tan` function is `sec^2`
   `d/dx(tan (xy))=sec^2 (xy)*(d (xy))/dx` (Chain Rule)
   `1+color(red)(sec^2(xy)(d(xy))/dx)=0`

4. Apply the Product Rule:
   `(d(xy))/dx=x dy/dx+y dx/dx=x dy/dx+ycancel(dx)/cancel(dx)`
   `1+sec^2 (xy)*color(red)((x dy/dx + y))=0`

5. Subtract 1 from both sides
   `sec^2 (xy)*(x dy/dx + y)=color(red)(-1)`

6. Divide both sides by `sec^2 (xy)`
   `x dy/dx + y = -1/color(red)(sec^2 (xy))`

7. `1/sec^2 (xy)=cos^2(xy)`
   `x dy/dx + y = -color(red)(cos^2 (xy))`

8. Subtract `y` from both sides
   `x dy/dx=-cos^2(xy)color(red)(-y)`

9. Divide both sides by `x`
   `dy/dx=(-cos^2(xy)-y)/color(red)(x)`

10. Simplify
    `dy/dx=color(red)(-) (cos^2(xy)color(red)(+)y)/x`