Question

Find the derivative implicity if x = tan xy. Help!?

Answer 1

Differentiate the expression.

`"d"/("d"x) (x) = "d"/("d"x) (tan(xy))`.

By the chain rule for the right hand side,

`1=sec^2(xy)*"d"/("d"x) (xy)`.

By the product rule,

`1=sec^2(xy)(y+x("d"y)/("d"x))`.

By the definition,

`x=tan(xy)`.

Then,

`x^2=tan^2(xy)`,
`x^2=-1+sec^2(xy)`,
`sec^2(xy)=x^2+1`.

Substituting,

`1=(1+x^2)(y+x("d"y)/("d"x))`,
`x("d"y)/("d"x) = 1/(1+x^2) - y`,
`("d"y)/("d"x) = 1/x (-y+1/(1+x^2))`.

Answer 2

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

Explanation:

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

However, we only differentiate explicit functions of `y` wrt `x`. But if we apply the chain rule we can differentiate an implicit function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

Example:

`d/dx(y^2) = d/dy(y^2)dy/dx = 2ydy/dx`

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

Now, we have:

`x=tan(xy)`

Implicitly differentiating wrt `x` (applying product rule):

`1 = sec^2(xy)d/dx(xy)`

`:. 1 = sec^2(xy)(xy' + y )`

`:. 1 = 1/cos^2(xy)(xy' + y )`

`:. cos^2(xy) = xy' + y`

`:. xy' = cos^2(xy) - y`

`:. y' = (cos^2(xy) - y)/x`

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) = tan(xy)-x`; Then;

`(partial F)/(partial x) = ysec^2(xy)-1`

`(partial F)/(partial y) = xsec^2(xy)`

And so:

`dy/dx = -(ysec^2(xy)-1)/(xsec^2(xy))`
`" " = -(y/cos^2(xy)-1)/(x/cos^2(xy)) * cos^2(xy)/cos^2(xy)`
`" " = -(y-cos^2(xy)) /x`
`" " = (cos^2(xy)-y) /x`, as before