Question

How do you use implicit differentiation to find the slope of the curve given `x^2y=x+2` at (2,1)?

Answer 1

`dy/dx=-3/4`

Explanation:

differentiate `color(blue)"implicitly with respect to x"`

The `color(blue)"product rule"` is required to differentiate `x^2y`

`x^2.dy/dx+2xy=1`

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

Use the point (2 ,1) to evaluate `dy/dx` which gives the slope.

`dy/dx=(1-(2xx2xx1))/2^2=-3/4`

Answer 2

Please see the explanation for the procedure. The slope is `-3/4`

Explanation:

I will explain each term.

Term 1: `(d[x^2y])/dx` requires the use of the product rule.

`(d[g(x)h(y)])/dx = g'(x)h(y) + g(x)h'(y)dy/dx`

`g(x) = x^2, g'(x) = 2x, h(y) = y, and h'(y) = 1`

`(d[x^2y])/dx = 2xy + x^2dy/dx`

Term 2:

`(d[x])/dx = 1`

Term 3:

`(d[2])/dx = 0`

Put these back into their corresponding locations in the equation:

`2xy + x^2dy/dx = 1 + 0`

Solve for `dy/dx`:

`x^2dy/dx = -2xy + 1`

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

The slope, m, is the above evaluated at the point `(2, 1)`

`m = (1 - 2(2)(1))/2^2`

`m = -3/4`