Question

How do you find `dy/dx` by implicit differentiation of `tan(x+y)=x` and evaluate at point (0,0)?

Answer 1

`dy/dx = [1-sec^2(x + y)]/sec^2(x + y)`

At `(0,0)`, `dy/dx = 0`

Explanation:

When doing implicit differentiation, you follow these essential steps:

1. Take the derivative of both sides of the equation with respect to `x`.
2. Differentiate terms with `x` as normal.
3. Differentiate terms with `y` as normal too but tag on a `dy/dx` to the end.
4. Solve for the `dy/dx`.

So, let's differentiate both sides:

`d/dx[tan(x + y)] = d/dx[x]`

The right hand side just comes out as `1`, but the left hand side will require that we use a chain rule. This goes as:

`d/dx[tan(x + y)] * d/dx[x + y]`

This comes out to:

`sec^2(x + y) * (1 + dy/dx)`

Putting this back in the whole equation:

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

Now, you just solve for `dy/dx` using some basic algebra:

`dy/dx = [1-sec^2(x + y)]/sec^2(x + y)`

Now, you're given the point `(0,0)` to evaluate the derivative at. All you do is plug this in:

`dy/dx = [1-sec^2(0)]/sec^2(0)`

`sec(0)` is simply `1/cos(0)`. Since `cos(0)` = 1, `sec(0)` is also 1. So, wherever we see `sec(0)`, we just plug in `1`:

`dy/dx = [1-1]/1 = 0`

So your tangent line would have a slope of `0` at the point `(0,0)`.

If you want more help in implicit differentiation, check out my video:

Hope that helped :)

Answer 2

`dy/dx = cos^2(x+y)-1`
Evaluating at `(0,0)` gives `dy/dx=0`

Explanation:

Implicit differentiation is just differentiation using chain rule.

`(d tan(x+y))/(d(x+y)) xx (d (x+y))/dx`=`dx/dx`
`sec^2(x+y)xx(1+dy/dx) = 1`

Rearranging:

`1+dy/dx = cos^2(x+y)`
`dy/dx = cos^2(x+y)-1`

Substituting `(0,0)` in above equation,
we get:
`dy/dx = cos^2(0)-1`
i.e `dy/dx=0`