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
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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` .
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