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The Power Rule can be proved using implicit differentiation for the case where $ n $ in a rational number, $ n = p/q, $ and $ y = f(x) = x" $ is assumed beforehand to be a differentiable.function. If $ y = x^{p/q}, $ then $ y^q = x^p. $ Use implicit differentiation to show that$ y' = \frac {p}{q} x^{(p/q)-1} $

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$$y^{\prime}=\left(\frac{p}{q}\right) \cdot x^{(p / q)-1}$$

01:13

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Derivatives

Differentiation

Campbell University

Oregon State University

Baylor University

University of Nottingham

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.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

04:06

The Power Rule can be prov…

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In this problem, we are asked to use simpson's differentiation to prove the power rule, so we're going to use this form less. Take derivative of both sides with respect to x. We have 2 times y times q minus 1 times y prime c y is the function of x is equal to p times x to the p mis 1 point. So from this we see that the y prime is of the form p times x times. P. Minus 1 divided by q times y q minus 1 on. So we can write this 1 as p over q times x to the p minus 1, divided by y to the cube minus 1 point now. We know why, in terms of x- and that is given right here, so we're just going to find out that we have then p over 2 times x to the p minus 1, divided by x to the p over q to the power of q. Minus 1. Point that is not equal to p over q times x to the power p minus 1 minus p times 2 plus p divided by 2. We can write this 1 as p over t times x to the p t minus 2 minus p, joule plus p divided by t these will get away and we then will have p over g times x to the p minus sorry p, over 2 minus 1 And we actually can write this 1 as p over q times x to the p over q minus 1 point, and we just put whatever is given in the problem state.

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