We have the expression:
[tex]-36x^{10}+9x^{36}y+y^3=-26[/tex]
And they ask for the differentiation of y with respect to x, then we need to assume based on the implicit derivate to y as y(x). Then we derivate both sides as:
[tex]\frac{dy}{dx}(-36x^{10}+9x^{36}y+y^3)=\frac{dy}{dx}(-26)[/tex][tex]-360x^9+324x^{35}y+9x^{36}\frac{\text{ dy}}{dx}+\frac{3y^2dy}{dx}=0[/tex]
We ordinate the expression and take factor comun the differentiation in order to leave it alone:
[tex]\frac{9x^{36}dy}{dx}+\frac{3y^2dy}{dx}=360x^9-324x^{35}y[/tex][tex]\frac{dy}{dx}(9x^{36}+3y^2)=360x^9-324x^{35}y[/tex][tex]\frac{dy}{dx}=\frac{360x^9-324x^{35}y}{9x^{35}+3y^2}=\frac{12x^9(10-9x^{26}y)}{3x^{35}+y^2}[/tex]
Then, we can say:
[tex]\frac{dy}{dx}=\frac{12x^9(10-9x^{26}y)}{3x^{35}+y^2}[/tex]
To obtain the tangent line we know that the slope of this tangent is the derivate evaluated in the given point, then:
[tex]m=\frac{dy}{dx}(1,1)=3[/tex]
And we use the standard formula for a line:
[tex]y-y_0=m(x-x_0)[/tex]
We reply our values:
[tex]y=3x-3+1[/tex][tex]y=3x-2[/tex]
Then the tangent line that pass through the curve in the point (1,1) is:
[tex]y=3x-2[/tex]
