$2x^6 + y^4 = 9xy$ at the point $(1, 2)$

Question

Find the equation of the tangent line to the curve $2x^6 + y^4 = 9xy$ at the point $(1, 2)$.

Can someone help me on this question, I am super confused and I have no clue where to start. I know the derivative of the equation is $12x^5 + 4y^3 = 9$ but after that I am lost. I plugged in $12*1+4*8$ but it is not $9$.

Please help

Answer 1

Let's start by assuming that $y$ is a function of $x$. This is important, as $\frac{d}{dx}y^n = ny^{n-1}y'$ by the chain rule. Thus, differentiating your equation, we get: $$\frac{d}{dx} (2x^6+y^4)=9\frac{d}{dx}(xy)$$ $$12x^5+4y^3y'=9y+9xy'$$ $$12x^5-9y=9xy'-4y^3y'$$ $$12x^5-9y=y'(9x-4y^3)$$ $$y'=\frac{12x^5-9y}{(9x-4y^3)}$$ Substituting $x=1$ and $y=2$ from the given point, we find that $$y'=\frac{6}{23}$$ All that is left is to apply point-slope form to find the equation of the tangent line $$y-y_1 = m(x-x_1)$$ $$y-2 = \frac{6}{23}(x-1)$$ $$y = \frac{6}{23}(x-1) + 2$$ $$y = \frac{6x}{23} + \frac{40}{23}$$

Answer 2

There is no such thing called "the derivative of the equation". You can not take "derivatives" of an equation; you take "derivatives" for functions.

The equation $2x^6+y^4=9xy$ implicitly define a function $y=f(x)$ near the point $(1,2)$. The derivative of this function $f$ at $x=1$ would give you the slope of the tangent line. Once you find out the derivative $f'(1)$, then the equation for the tangent line would be $$ y-2=f'(1)(x-1)\;\tag{0} $$

In order the find $f'(1)$, you use implicit differentiation as follows: $$ 12x^5+4y^3y'=9(y+xy')\tag{1} $$ where $y'=f'(x)$. Solving (1), you get $$ y'=\frac{12x^5-9y}{9x-4y^3}\tag{2} $$ Substituting $(x,y)=(1,2)$ into (2), you get the value of $f'(1)$.