find the equation of the tangent line at the given value of x on the curve. 4y³(x - 6)+x√y = 12; x = 6 y = □

Explanation:

Step1: Find the y - value when x = 6

Substitute x = 6 into the equation $4y^{3}(x - 6)+x\sqrt{y}=12$. When x = 6, the first - term $4y^{3}(x - 6)=4y^{3}(6 - 6)=0$. The equation becomes $6\sqrt{y}=12$. Solve for y: $\sqrt{y}=2$, so $y = 4$.

Step2: Differentiate the given equation implicitly with respect to x

Using the product rule and chain rule: The derivative of $4y^{3}(x - 6)$ with respect to x is $4y^{3}+12y^{2}(x - 6)y'$ (product rule: $(uv)'=u'v + uv'$, where $u = 4y^{3}$ and $v=x - 6$). The derivative of $x\sqrt{y}$ with respect to x is $\sqrt{y}+\frac{x}{2\sqrt{y}}y'$ (product rule: $u = x$, $v=\sqrt{y}$). The derivative of the right - hand side 12 with respect to x is 0. So, $4y^{3}+12y^{2}(x - 6)y'+\sqrt{y}+\frac{x}{2\sqrt{y}}y'=0$.

Step3: Substitute x = 6 and y = 4 into the derivative equation

When x = 6 and y = 4, the equation $4y^{3}+12y^{2}(x - 6)y'+\sqrt{y}+\frac{x}{2\sqrt{y}}y'=0$ becomes: $4\times4^{3}+\sqrt{4}+\frac{6}{2\sqrt{4}}y'=0$. $4\times64 + 2+\frac{6}{4}y'=0$. $256+2+\frac{3}{2}y'=0$. $258+\frac{3}{2}y'=0$. $\frac{3}{2}y'=-258$. $y'=-172$.

Step4: Use the point - slope form of a line to find the tangent line equation

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(6,4)$ and $m=-172$. $y - 4=-172(x - 6)$. $y-4=-172x + 1032$. $y=-172x+1036$.

Answer:

$y=-172x + 1036$