Question

Given the equation below , find (dy)/(dx); - 36x ^ 10 + 9x ^ 36 * y + y ^ 3 = - 26

Answer 1

We have the expression:

$-36x^{10}+9x^{36}y+y^3=-26$

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:

$\frac{dy}{dx}(-36x^{10}+9x^{36}y+y^3)=\frac{dy}{dx}(-26)$$-360x^9+324x^{35}y+9x^{36}\frac{\text{ dy}}{dx}+\frac{3y^2dy}{dx}=0$

We ordinate the expression and take factor comun the differentiation in order to leave it alone:

$\frac{9x^{36}dy}{dx}+\frac{3y^2dy}{dx}=360x^9-324x^{35}y$$\frac{dy}{dx}(9x^{36}+3y^2)=360x^9-324x^{35}y$$\frac{dy}{dx}=\frac{360x^9-324x^{35}y}{9x^{35}+3y^2}=\frac{12x^9(10-9x^{26}y)}{3x^{35}+y^2}$

Then, we can say:

$\frac{dy}{dx}=\frac{12x^9(10-9x^{26}y)}{3x^{35}+y^2}$

To obtain the tangent line we know that the slope of this tangent is the derivate evaluated in the given point, then:

$m=\frac{dy}{dx}(1,1)=3$

And we use the standard formula for a line:

$y-y_0=m(x-x_0)$

We reply our values:

$y=3x-3+1$$y=3x-2$

Then the tangent line that pass through the curve in the point (1,1) is:

$y=3x-2$