Find y'' by implicit differentiation.7x2 + y2 = 5

Sure, let's find the second derivative of the given equation using implicit differentiation.

The given equation is 7x² + y² = 5.

Step 1: Differentiate both sides of the equation with respect to x.

The derivative of 7x² with respect to x is 14x. The derivative of y² with respect to x is 2y * dy/dx (using the chain rule). The derivative of 5 with respect to x is 0.

So, we get 14x + 2y * dy/dx = 0.

Step 2: Solve the above equation for dy/dx.

dy/dx = -14x / (2y) = -7x/y.

This is the first derivative.

Step 3: Differentiate dy/dx with respect to x to get the second derivative.

The derivative of -7x/y with respect to x is a bit tricky because it's a quotient. We'll use the quotient rule, which states that the derivative of u/v is (v * du/dx - u * dv/dx) / v².

Here, u = -7x and v = y. So, du/dx = -7 and dv/dx = dy/dx.

Substituting these into the quotient rule gives us:

y'' = (y * -7 - -7x * dy/dx) / y² = (-7y + 7x * dy/dx) / y².

But we already know that dy/dx = -7x/y. Substituting this into the equation gives us:

y'' = (-7y + 7x * -7x/y) / y² = (-7y - 49x²/y) / y² = -7/y - 49x²/y³.

So, the second derivative of the given equation is y'' = -7/y - 49x²/y³.