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)
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
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+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.
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).
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.
Plug-in m=7, x1=1 and y1= 0.
`y=7x - 7`
Therefore, the equation of the tangent line is `y = 7x - 7` .