Differentiation by the chain rule

jtt
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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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