yah, implicit differentiation or
something
differentiate with respect to x
take deririvitve
ofboth sides
[tex]2x+2y \frac{dy}{dx} =0[/tex]
solve for [tex] \frac{dy}{dx}[/tex]
minus 2x both sides
[tex]2y \frac{dy}{dx} =-2x[/tex]
divide both sides by 2y
[tex] \frac{dy}{dx} =\frac{-x}{y}[/tex]
now, take the derivitive of that
note that [tex] y =\frac{-x}{y}[/tex]
[tex] \frac{(-x)'(y)-(y')(-x)}{y^2} [/tex]
sub [tex] \frac{-x}{y}[/tex] for y
[tex] \frac{(-y)-(\frac{-x}{y})(-x)}{y^2} [/tex]
[tex] \frac{-y-\frac{x^2}{y}}{y^2} [/tex]
Applying implicit differentiation, it is found that:
[tex]\frac{d^2y}{dx^2}(4,3) = \frac{7}{27}[/tex]
The curve given is:
[tex]x^2 + y^2 = 25[/tex]
Applying implicit differentiation, we have that the first derivative is:
[tex]2x\frac{dx}{dx} + 2y\frac{dy}{dx} = 0[/tex]
[tex]2y\frac{dy}{dx} = -2x[/tex]
[tex]\frac{dy}{dx} = -\frac{x}{y}[/tex]
Again applying implicit differentiation, the second derivative is:
[tex]\frac{d^2y}{dx^2} = -\frac{y\frac{dx}{dx} - x\frac{dy}{dx}}{y^2}[/tex]
Since [tex]\frac{dy}{dx} = -\frac{x}{y}[/tex]
[tex]\frac{d^2y}{dx^2} = -\frac{y - \frac{x^2}{y}}{y^2}[/tex]
[tex]\frac{d^2y}{dx^2} = \frac{x^2 - y^2}{y^3}[/tex]
At point (4,3), [tex]x = 4, y = 3[/tex], then:
[tex]\frac{d^2y}{dx^2}(4,3) = \frac{4^2 - 3^2}{3^3} = \frac{7}{27}[/tex]
A similar problem is given at https://brainly.com/question/15278071
