I am supposed to determine the equation for the tangent in point (4,1) to the curve:
$$5\sqrt{x}=2\sqrt{y}(x+y^2)$$
I think that I should differentiate the expression and then put the values (4,1) where x and y are. But how do I differentiate this right when both x and y are involved?
You can differentiate RHS as the following : $$\begin{align}\frac{d}{dx}(2\sqrt y\ (x+y^2))&=\left(\frac{d}{dx}(2\sqrt y)\right)(x+y^2)+2\sqrt y\left(\frac{d}{dx}(x+y^2)\right)\\&=\left(\frac{d(2\sqrt y)}{dy}\cdot\frac{dy}{dx}\right)(x+y^2)+2\sqrt y\ \left(1+\frac{d(y^2)}{dy}\cdot\frac{dy}{dx}\right)\\&=\left(\frac{1}{\sqrt y}\cdot y'\right)(x+y^2)+2\sqrt y\ (1+2yy').\end{align}$$
So, since you'll have $$\frac{5}{2\sqrt x}=\left(\frac{1}{\sqrt y}\cdot y'\right)(x+y^2)+2\sqrt y\ (1+2yy'),$$ you'll have the following to solove for $y'$ : $$\frac{5}{2\sqrt 4}=\left(\frac{1}{\sqrt 1}\cdot y'\right)(4+1^2)+2\sqrt 1\ (1+2\cdot 1\cdot y').$$ The answer will be $y-1=y'(x-4)$.
Use implicit differentiation, let's define $F(x,y)$ as follows: $$F(x,y)=5\sqrt{x}-2\sqrt{y}x-2\sqrt{y}y^2$$ We know from implicit differentiation that $$\frac{dy}{dx} = -\frac {F_x}{F_y}.$$ So by calculating $-\frac {F_x}{F_y}$ we obtain $$\frac{dy}{dx} =\frac{(5-4\sqrt{xy})\sqrt{y}}{2\sqrt{x}(x+5y^2)}$$ Plugging the point $(4,1)$ at $\frac{dy}{dx}$ we find the slope of tangent line as $-1/12$. Finally the equation of the tangent line is $$y-1=-\frac{1}{12}(x-4)$$
