What is the second derivative of x2+y2=25 evaluated at (-3, -4)?

Question

eyakhiwabhn · Accepted Answer

We can treat y as a constant if we want to find the derivative with respect to x. (derivative of a constant is zero).

\frac{d}{d x} (x^{2} + y^{2} = 25)

The sum rule allows us to divide this up

{(a + b)}^{'} = a^{'} + b^{'}

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2} = 25)

Using the power rule,

\frac{d}{d x} (x^{2})

becomes 2x, and if we treat

y^{2}

as a constant, the derivative of that and 25 becomes 0. We're just left with 2x.

\frac{d}{d x} = 2 x

Finding the Second Derivative:

\frac{d}{d x} (2 x) = 2

We arrive at 2 by finding the second derivative.

John Noble · Accepted Answer

Given:x2+y2=25This should be recognized as a circle with a radius of 5 and a center at the origin.Differentiating Implicitly wrt #x# we get:2x+2ydydx=0∴ydydx=−x∴dydx=−xyNow, using the quotient rule, we differentiate a second time (implicitly):d2ydx2=−(y)(1)−(x)(dydx)(y)2=−y−(x)(−xy)y2=−y+x2yy2=−y2+x2yy2=−y2+x2y3=−25y3So at the point (-3,-4) we have:d2ydx2=−25(−4)3=−25−64=2564

What is the second derivative of x^2+y^2=25 evaluated at (-3, -4)?

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