Now, take the derivative of y'.ObviousManiac said:Homework Statement
Find y" in terms of x and y:
y^2 + 2y = 2x + 1
Homework Equations
N/A
The Attempt at a Solution
I found the first derivative:
y^2 + 2y = 2x + 1
2yy'+2y'=2
2y'.(y+1)=2
y'=2/2(y+1)
y'=1/(y+1)
But I'm having trouble moving on from there.
SammyS said:Now, take the derivative of y'.
Sure it will have y' in it, but then substitute the result you have for y' into that.
The second derivative in terms of x and y is a measure of how the rate of change of the first derivative changes with respect to changes in both x and y. It can be calculated by taking the derivative of the first derivative, or by using the chain rule for multivariable functions.
The second derivative can be used to determine the concavity of a function. If the second derivative is positive at a point, the function is concave up at that point. If the second derivative is negative, the function is concave down. The points where the second derivative changes sign (from positive to negative or vice versa) are called inflection points.
Yes, it is possible for the second derivative to be negative even if the first derivative is positive. This means that the function is increasing at a decreasing rate. In other words, the slope of the tangent line is getting smaller as x increases.
The second derivative can be used in optimization problems to determine whether a critical point (where the first derivative equals 0) is a maximum or minimum. If the second derivative is positive at a critical point, it is a minimum. If the second derivative is negative, it is a maximum.
No, the second derivative itself does not directly represent the rate of change of a function. However, it can be used to calculate the rate of change of the first derivative, which represents the rate of change of the original function. This can be useful in certain applications, such as in physics where the second derivative of position with respect to time represents acceleration.