# Use implicit differentiation to find an equation of the tangent line to the graph at the given point. `x+y-1=ln(x^8+y^11)` , (1,0) y=_____? I know we must find the derivative, but I'm not sure...

Use implicit differentiation to find an equation of the tangent line to the graph at the given point.

`x+y-1=ln(x^8+y^11)` , (1,0)

y=_____?

I know we must find the derivative, but I'm not sure to continue. I was given an example earlier. but it was wrong and left me clueless, if anyone can help, it would be appreciated

### 1 Answer | Add Yours

`x+y-1=ln(x^8+y^11)`

To solve this, we have to determine the slope of the tangent line. Since

Slope of Tangent, m `= dy/dx`

then, we have to apply implicit differentiation.

So, take the derivative of each side of the equation with respect to x.

`d/dx (x+y-1)=d/dx (ln(x^8+y^11))`

`1+dy/dx - 0 = 1/(x^8+y^11)*d/dx(x^8+y^11)`

`1+dy/dx = 1/(x^8+y^11)*(8x^7+11y^10dy/dx)`

Then, isolate dy/dx. To do so, multiply both sides by the denominator.

`(x^8+y^11)*(1+dy/dx)=1/(x^8+y^11)(8x^7+11y^10dy/dx)*(x^8+y^11)`

`x^8+y^11+x^8dy/dx+y^11dy/dx = 8x^7+11y^10dy/dx`

Then, bring together the terms with dy/dx on one side of the equation. And bring together the terms without dy/dx to the other side.

`x^8dy/dx + y^11dy/dx - 11y^10dy/dx = 8x^7-x^8-y^11`

`x^8dy/dx + y^11dy/dx - 11y^10dy/dx = -x^8+8x^7-y^11`

Factor out the GCF of the left side of the equation.

`dy/dx(x^8+y^11-11y^10)=-x^8+8x^7-y^11`

And, divide both sides by x^8+y^11-11y^10.

`dy/dx = (-x^8+8x^7-y^11)/(x^8+y^11-11y^10)`

Now that the dy/dx of the equation is known, plug-in the point of tangency which is (1,0).

`dy/dx= (-1^8+8(1)^7-0^11)/(1^8+0^11-11(0)^10)`

`dy/dx=(-1+8-0)/(1+0-0)`

`dy/dx=7`

`m =7`

Hence, the tangent line has a slope of 7 and passes the point (1,0).

Next, to determine its equation, apply the point-slope form.

`y-y_1=m(x-x_1)`

Plug-in m=7, x1=1 and y1= 0.

`y-0=7(x-1)`

`y=7x - 7`

**Therefore, the equation of the tangent line is `y = 7x - 7` .**