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Find the second derivative at (4,3)X^2+y^2=25

Sagot :

Given the equation:

[tex]x^2+y^2=25[/tex]

Let's find the second derivative at the point (4, 3).

Subtract 25 from both sides to equate to zero

[tex]\begin{gathered} x^2+y^2-25=25-25 \\ \\ x^2+y^2-25=0 \end{gathered}[/tex]

To find the second derivative, first find the first derivative:

[tex]\begin{gathered} 2x+2y\frac{dy}{dx}=0 \\ \\ \frac{dy}{dx}=-\frac{x}{y} \end{gathered}[/tex]

Now find the second derivative:

[tex]\begin{gathered} \frac{d^2y}{dx^2}=-\frac{x^{\prime}(y)-xy^{\prime}}{y^2} \\ \\ \frac{d^2y}{dx^2}=-\frac{y-x\frac{dy}{dx}}{y^2} \end{gathered}[/tex]

Now, pulg in 4 for x and 3 for y:

dy/dx = -4/3

[tex]\begin{gathered} \frac{d^2y}{dx^2}=-\frac{3-4(-\frac{4}{3})}{3^2} \\ \\ \frac{d^{2}y}{dx^{2}}=-\frac{9+16}{27}=-\frac{25}{27} \end{gathered}[/tex]

ANSWER:

[tex]-\frac{25}{27}[/tex]