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If $ g $ is a twice differentiable function and $ f(x) = xg(x^2), $ find $ f" $ in terms of $ g, g', $ and $ g". $

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$f^{\prime \prime}(x)=6 x g^{\prime}\left(x^{2}\right)+4 x^{3} g^{\prime \prime}\left(x^{2}\right)$

01:13

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Derivatives

Differentiation

Campbell University

Baylor University

University of Michigan - Ann Arbor

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.

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in this problem, we're going to find F double prime of X. So of course we have to find f prime first. So the first derivative would be using the product rule. We have the first X times, the derivative of a second, and to find the derivative of G of X squared, we need to use the chain rule. So the derivative of the outside would be g prime of X squared times the derivative of the inside two x. So so far we have half the product rule done. We have the first times a derivative of the second. Now we have plus the second G of X squared times, the derivative of the first, which is just one the derivative of X is one. Okay, let's simplify that a little bit. So we have two x squared times g prime of X squared plus g of X squared, and we're going to find the derivative of this and that will be our second derivative. So f double prime notice. We have a product again. So we're going to use the product rule on that. So we have the 1st 2 x squared times, the derivative of the second. So the derivative of J Prime would be G double prime of X squared times, the derivative of the inside two x plus. The second g prime of X squared times a derivative of the first, The derivative of two X squared would be four x So all of this was the product rule on the first part. Now we still have plus the derivative of the second part, and the derivative of G would be g prime of X squared times, a derivative of the inside two x. Okay, let's see what we can do to simplify. So in the first term, we have two x squared times two x will multiply those and that's four x cubed with four x cubed times G double prime of X squared and then our second term is four x 10 g prime and our third term is two x times g prime. So we can add the forex and the two X and we get six x times g prime of X squared. Now the problem says we can give our answer in terms of G prime and G so we can leave it like that

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