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$$\text { Find } \frac{d^{2} y}{d x^{2}}, \text { if } 3 x^{2}-2 y^{2}=6$$

$$\frac{-9}{2 y^{3}}$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 3

Concavity and the Second Derivative

Derivatives

Campbell University

Oregon State University

Harvey Mudd College

Baylor University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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we want to find D squared Y D x squared for three X squared minus two. Y squared equals six. To do so we're going to use implicit differentiation. That is we're gonna differentiate every time with respect to X. Where every term including why we'll have a differential dy dx as a consequence of the chain rule. So by implicit differentiation six x minus for y dy dx equals zero. Thus we solve for dy dx as dy dx equals 3/2 X Y or three half X Y negative. First we have to differentiate again, again applying implicit differentiation to solve so implicit differentiation repeating gives the square Y. D X squared equals 3/2 Y minus three half X. Y and second dy dx. This is by the product rule. Thus we plug in our G Y. D. Action before to obtain a solution D squared Y. The X squared equals three half over y minus 9/4 expired over white cube. Notice how here as I mentioned, we had to plug in our dy dx from above. That is we apply both chain rule and the product school

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