# Find y' and the slope of the tangent line to the graph of the equation: 6sq rt.y^+1-2x^3/2-2=0 at the point (4,3) The sq rt symbol stops at the y^3+1

### 1 Answer | Add Yours

`6sqrt(y^3+1)-2x^(3/2) - 2 = 0`

Using implicit differentiation,

`6 xx1/2 (y^3+1)^(-1/2) xx 3y^2 xxdy - 2 xx 3/2 xx x^(1/2)dx - 0 = 0`

`(9y^2)/sqrt(y^3+1)xxdy- 3sqrt(x)dx = 0`

`(dy)/(dx) = (3sqrt(x))/((9y^2)/sqrt(y^3+1))`

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

At (4,3),

`(dy)/(dx) = (sqrt(4)sqrt(3^3+1))/(3xx3^2)`

`(dy)/(dx) = (2sqrt(10))/(27)`

**Therefore the slope of the tangent line to the graph at (4,3) is** `(2sqrt(10))/(27)`