Differentiation by the chain rule

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Homework Help Overview

The discussion revolves around finding the derivative of the function Y = x^3(5x-1)^4, focusing on the application of differentiation rules, particularly the product rule and the chain rule.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of the product rule and the chain rule, with some expressing confusion about the correct steps to take. There are attempts to clarify the roles of the functions involved and the necessary derivatives.

Discussion Status

Some participants have provided guidance on the need to apply the product rule first, while others have attempted to clarify the use of the chain rule in the context of the problem. Multiple interpretations of the approach are being discussed, but there is no explicit consensus on the correct method yet.

Contextual Notes

Participants are navigating through the differentiation process and expressing confusion regarding the application of the rules, particularly in the context of the function's structure. There is an emphasis on ensuring the correct derivatives are found for both components of the product.

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Homework Statement


Find the derivative of the following:


Homework Equations


Y= x^3(5x-1)^4


The Attempt at a Solution


4(3x^2(5x-1)^3)(4(3x^2(3(5x-1)^2)(2(5x-1)(5)
 
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That doesn't look like the chain rule to me. Apply the product rule first.
 
i tried bringing down the 4Th exponent and then subtract it by one to get three, then leaving the inside alone ( 5x-1) at the same time taking the derivative of 3x^2. after that i got confused and got a wrong answer.
 
y = f(x)g(x) where f(x)= x^3 and g(x)=(5x-1)^4
So you'll first need to apply the product rule... as you do you'll need the derivative of g.

g(x) = P\circ L (x) = P( L(x)) where P(x) = x^4 and L(x)=5x - 1. As a composition you need to apply the chain rule. (P for power, L for linear).

If you'd rather use the Leibniz notation form of the chain rule: \frac{du}{dx} = \frac{du}{dv} \frac{dv}{dx} then let u=g(x) = P(v) with v = L(x).
 
Your function is f*g where f=x^3 and g=(5x-1)^4, right? The product rule says the derivative of f*g is f'*g+f*g', also right? Now you just need to find f' and g'. Finding the derivative of g' is where you need the chain rule.
 

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