Question

What is the implicit derivative of `5=x-1/(xy^2)`?

Answer 1

See below.

Explanation:

When finding the implicit derivative of an equation, we are finding the derivative in respect to `x`.

This means that when we find the derivative of any variable other than `x`, we need to tag on a `(dsquare)/dx` `(square "is the variable")` `"to the end".`

Knowing this, we will start with implicit differentiation.

Here is what we are given:

`5=x-1/(xy^2)`

We can go ahead and avoid the Quotient Rule by multiply each term by `xy^2`.

`5=x-1/(xy^2)`

`=> 5xy^2=x^2y^2-1`

Here we can use the Product Rule to solve:

`5xy^2=x^2y^2-1`

`=> 5(y^2)+5x(2ydy/dx)=2x(y^2)+x^2(2ydy/dx)`

Simplify:

`=> 5y^2+10xydy/dx=2xy^2+2x^2ydy/dx`

Solve for `dy/dx`:

`=> 10xydy/dx-2x^2ydy/dx=2xy^2-5y^2`

`=> dy/dx(10xy-2x^2y)=2xy^2-5y^2`

`=>dy/dx=(2xy^2-5y^2)/(10xy-2x^2y)`

`=>dy/dx=(2xy-5y)/(10x-2x^2)`

Answer 2

Please see below.

Explanation:

I assume that we want `dy/dx` given that

`5 = x-1/(xy^2`

Method 1

To avoid the quotient rule, I would start by multiplying both sides be `xy^2` to get

`5xy^2 = x^2y^2-1`

Differentiating with respect to `x` (and using the product rule) gets us

`d/dx(5xy^2) = d/dx(x^2y^2-1)`

`(5)y^2+5x(2ydy/dx) = (2x)y^2+x^2(2ydy/dx) - 0`

Now we'll do some algebra.

`5y^2+10xydy/dx = 2xy^2+2x^2ydy/dx`

`10xydy/dx - 2x^2ydy/dx= 2xy^2- 5y^2`

`dy/dx = ( 2xy^2- 5y^2)/(10xy-2x^2y)`

`= ( 2xy - 5y)/(10x - 2x^2)`

Method 2

`d/dx(5) = d/dx(x-1/(xy^2))`

`0 = 1+(1y^2+2xy(dy/dx))/(x^2y^4)`

`0 = 1+(y+2x(dy/dx))/(x^2y^3)`

`0 = x^2y^3+y+2x dy/dx`

`dy/dx = - (x^2y^3+y)/(2x)`

Resolution

To see that these are equivalent, start with

`dy/dx = ( 2xy - 5y)/(10x - 2x^2)`

`= ( 2xy - (5)y)/(2x(5) - 2x^2)`

Now replace `5` by `x-1/(xy^2)`

`= ( 2xy - (x-1/(xy^2))y)/(2x(x-1/(xy^2)) - 2x^2)`

Simplify algebraically to get

`= - (x^2y^3+y)/(2x)`