# Use implicit differentiation to find an equation of the tangent line to the graph at the given point. x+y-1=ln(x^12+y^4) y=____? Any Help Please? my book or notes do not have any similar problems

*print*Print*list*Cite

### 2 Answers

We need to differentiate both sides of the equation.

We will apply the Power rule of Differentiation:

`(d)/(dx)(x^n) = nx^(n - 1)`

So, for the derivative of x it will be:

`(d)/(dx)(x) = 1x^(1-1) = 1x^0 = 1.`

We will also apply the formula for the derivative of ln.

`(d)/(dx)(lnu) = (du)/(u)`

So, for the derivative of the right side, we will have:

`(d)/(dx)(ln(x^12+y^4)) = (12x^11 + 4y^3*y')/(x^12 + y^4)`

Take note that` (d)/(dx)(x^12) = 12x^(12-1) = 12x^11` .

Same thing with the derivative of` y^4` , it will be `(d)/(dx)(y^4) = 4y^(4-1)*y' = 4y^3*y'` .

We multiply it by y', since the y is the dependent variable, we do it when we are taking the implicit differentiation.

So, the equation will be:

`1 + y' = (12x^11 + 4y^3*y')/(x^12 + y^4)`

Multiply both sides by `(x^12 + y^4)` .

`(1 + y') (x^12 + y^4) = (12x^11 + 4y^3*y')/(x^12 + y^4)* (x^12 + y^4)`

Apply foil method on left side. On right side, cancel out common factor on top and bottom.

`(1 + y') (x^12 + y^4) = (1*x^12) + (1*y^4) + (y'*x^12) +(y' *y^4) = x^12 +y^4 +x^12y' +y^4y'`

`(12x^11 + 4y^3*y')/(x^12 + y^4)* (x^12 + y^4) = 12x^11 + 4y^3*y'`

So, we will have:

`x^12 +y^4 +x^12y' +y^4y' = 12x^11 + 4y^3*y'`

Put all the terms with `y'` on left side and all the other terms on right side.

Remember to change the sign of the term that you will be putting to the other side of the equation.

`x^12y' +y^4y' - 4y^3*y' = 12x^11 - x^12 - y^4 `

Factor out` y'` on terms on left side.

`y'(x^12 +y^4 - 4y^3) = 12x^11 - x^12 - y^4 `

Isolate the ` y' ` on left side, divide both sides by `(x^12 +y^4 - 4y^3)` .

`y' = (12x^11 - x^12 - y^4)/(x^12 +y^4 - 4y^3)`

Since we are given with a point `(1, 0)` , we plug-in` x = 1 ` and `y = 0` .

`y' = (12(1)^11 - (1)^12 - (1)^4)/((1)^12 +(0)^4 - 4(0)^3)`

`= (12 - 1 - 1)/(1 + 0 - 0)`

`y' = 12`

Hence, the slope of the tangent line is `12` .

Apply the two point form: `y - y_1 = m(x - x_1)` .

`y - 0 = 12(x - 1)`

Use the Distributive property on right side `a(b + c) = a*b + a*` c.

`y = 12(x - 1) `

`y = 12x - 12`

That is it!

The Given Point is (1,0) I forgot to include that.