Question

What is the slope of `xy^3+xy=10` at the point `(5,1)`?

Answer 1

Slope `=-1/10` at `(5, 1)`

Explanation:

`xy^3+xy=10`

Applying implicit differentiation:

`x*3y^2dy/dx+y^3*1 +xdy/dx+y*1 =0`

`(3xy^2+x)dy/dx = -(y^3+y)`

`:. dy/dx = (-(y^3+y))/(3xy^2+x)`

`dy/dx` represents the slope of the tangent to the curve at any point `(x_1, y_1)` on the curve.

Slope of the tangent `(m)` at `(5, 1)`

`= (-(1+1))/(3*5*1 +5) = (-2)/(15+5) = -2/20`

`=-1/10`

Answer 2

`"slope "=-1/10`

Explanation:

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

`"differentiate "xy^3" and "xy" using the "color(blue)"product rule"`

#(x.3y^2dy/dx+y^3 .1)+(xdy/dx+y.1)=0#

`rArr3xy^2dy/dx+y^3+xdy/dx+y=0`

`rArrdy/dx(3xy^2+x)=-y^3-y`

`rArrdy/dx=-(y^3+y)/(3xy^2+x)`

`"at "(5,1)`

`dy/dx=-(1+1)/(15+5)=-2/20=-1/10larrcolor(red)"slope"`