Question

Find the derivative of `x^2 + y^2 = (5x^2 + 4y^2 -x)^2`, and find the equation of the tangent at `(0,0.25)`?

Answer 1

The derivative is given (implicitly) by;

`x + ydy/dx = (5x^2 + 4y^2 -x)(10x+8ydy/dx-1)`

The equation of the tangent at `(0,0.25)` is:

`y = x + 1/4`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

When we differentiate `y` wrt `x` we get `dy/dx`.

However, we cannot differentiate a non implicit function of `y` wrt `x`. But if we apply the chain rule we can differentiate a function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

When this is done in situ it is known as implicit differentiation.

We have:

`x^2 + y^2 = (5x^2 + 4y^2 -x)^2`

Differentiate wrt `x` (applying chain rule):

`2x+2ydy/dx = 2(5x^2 + 4y^2 -x)(10x+8ydy/dx-1)`

`:. x + ydy/dx = (5x^2 + 4y^2 -x)(10x+8ydy/dx-1)`

Whilst we could spend time and effort to find an explicit equation for `dy/dx`, we have no requirement to, so lets just find `dy/dx` at the point `(0,0.25)`;

`=> 0 + 1/4 dy/dx = (0 + 4(1/16) -0)(0+8/4dy/dx-1)`
`:. \ \ \ \ \ \ \ \ 1/4 dy/dx = (1/4)(2dy/dx-1)`
`:. \ \ \ \ \ \ \ \ 1/4 dy/dx = 1/2dy/dx-1/4`
`:. \ \ \ \ \ \ \ \ 1/4 dy/dx = 1/4`

`:. \ \ \ \ \ \ \ \ \ \ \ \ dy/dx = 1`

So the tangent passes through `(0,0.25)` and has gradient `1` so using the point/slope form `y-y_1=m(x-x_1)` the equation we seek is;

`\ \ \ \ \ y-1/4 = 1(x-0)`
`:. y-1/4 = x`
`:. \ \ \ \ \ \ \ \ y = x + 1/4`

We can verify this solution graphically;

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

So Let `F(x,y) = x^2 + y^2 - (5x^2 + 4y^2 -x)^2`; Then;

`(partial F)/(partial x) = 2x - 2(5x^2 + 4y^2 -x)(10x-1)`

`(partial F)/(partial y) = 2y - 2(5x^2 + 4y^2 -x)(8y)`

And so:

`dy/dx = -(2x - 2(5x^2 + 4y^2 -x)(10x-1))/(2y - 2(5x^2 + 4y^2 -x)(8y))`
`\ \ \ \ \ \= -(x - (5x^2 + 4y^2 -x)(10x-1))/(y - 8y(5x^2 + 4y^2 -x))`

Here we get an immediate implicit function for the derivative, so again at `(0,0.25)` we have:

`dy/dx = -(0 - (0 + 4/16 -0)(0-1))/(1/4 - 8/4(0 + 4/16 -0))`
`\ \ \ \ \ \ = -(- (1/4)(-1))/(1/4 - 2(1/4))`
`\ \ \ \ \ \ = -(1/4)/(-1/4)`
`\ \ \ \ \ \ = 1 \ \ \ \`, as before