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If x² . y² = 25, what is the value of (d² .y/dx²) at the point (4,3)?

A. 0
B. 3/4
C. -4/3
D. -3/4

1 Answer

5 votes

Final answer:

The value of (d²y/dx²) at the point (4,3) is determined by differentiating the equation x² · y² = 25 twice with respect to x. The first differentiation, using the product rule, reveals that (dy/dx) = 0. Subsequently, the second derivative (d²y/dx²) is also 0, which corresponds to answer choice A.

Step-by-step explanation:

The question asks for the value of (d²y/dx²) at a given point when x² · y² = 25. To find this value, we need to differentiate the given equation with respect to x, twice. Starting by differentiating both sides with respect to x:

First differentiation (using the product rule):2xy(dy/dx) + 2x(dy/dx)y = 0.

Now, solve for dy/dx and then differentiate this result with respect to x to find d²y/dx².

At the point (4,3), we have:

2(4)(3)(dy/dx) + 2(4)(dy/dx)(3) = 0,

24(dy/dx) + 24(dy/dx) = 0,

48(dy/dx) = 0,

(dy/dx) = 0 (since 48 is not equal to zero).

Second differentiation:

Since (dy/dx) = 0, it implies that all terms in the second derivative that contain (dy/dx) will be zero, therefore (d²y/dx²) = 0 at the point (4,3).

The correct answer is A. 0.

User Jacole
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