I keep failing to understand what I'm doing wrong - derivatives

Question

I need to find the derivative of this function:

$f(x)=x^2(x-2)^4$

I have gone about this two ways:

1. Chain Rule inside a Power Rule

$$f(x)=x^2(x-2)^4$$ (I found the derivative of $(x-2)^4$ using the chain rule.) $$f'(x)=(x^2)(4)(x-2)^3(1)+(x-2)^4(2x)$$ $$f'(x)=4x^2(x-2)^3+2x(x-2)^4$$

2. Power Rule only

$$f(x)=x^2(x-2)^4$$ $$f'(x)=(x^2)(4(x-2)^3)+(x-2)^4(2x)$$ $$f'(x)=4x^2(x-2)^3+2x(x-2)^4$$

My calculator gives me the same answer that I get in method 2. However the textbook says the answer to this problem is $2x(x-2)^3(3x-2)$

So, assuming the textbook is correct (it better be), what are my calculator and I doing wrong? Also, why does the using the chain rule inside of the power rule produce a different result? I only ask the second question because I was under the impression that the chain rule was just another shortcut for finding derivatives.

Answer 1

Both terms in $4x^2(x-2)^3+2x(x-2)^4$ are multiples of $2x$ and $(x-2)^3$. So factor them out: $$ 4x^2(x-2)^3+2x(x-2)^4 = 2x(x-2)^3[2x+(x-2)] = 2x(x-2)^3(3x-2) $$

Answer 2

f(x) = X²(x-2)^4

In this question, U will have to apply products rule and Chain rule

Let U = x² V= (x-2)^4

du/dx = 2x

But, V is a function of function, so to differentiate This, Let x-2 = g

So, V=g^4

dg/dx = 1

dv/dg = 4g³

So, dv/dx =dv/dg × dg/dx

So, dv/dx = 4g³ ×1

Since g= x-2

dv/dx = 4(x-2)³

So back to our dy/dx

dy/dx = Udv/dx + Vdu/dx

Our U=x², V = (x-2)^4

du/dx = 2x , dv/dx = 4(x-2)³

dy/dx = x² (4(x-2)³) + (x-2)^4(2x)

dy/dx = 4x²(x-2)³ + 2x(x-2)^4

Factorise

dy/dx = 2x(x-2)³ (2x +(x-2)

dy/dx = 2x(x-2)³(3x-2)

So, ur textbook is correct, and u are also correct. Your textbook only simplified further by factorising.

And, as for ur second question, Each formula in differentiation has its uses, Quotient Rule for quotient, Product rule for product questions and Chain Rule for Function of a function question, u can't interchange them.