My Math Forum Derivative of (x^2-1)^2 problem

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 February 23rd, 2019, 10:58 PM #1 Newbie     Joined: Feb 2019 From: N/A Posts: 18 Thanks: 0 Math Focus: Calculus Derivative of (x^2-1)^2 problem Is the answer: (2x^2-2)?
 February 24th, 2019, 06:47 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,393 Thanks: 749 Chain rule.
 February 24th, 2019, 08:54 AM #3 Senior Member     Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics Use the chain rule to solve this. Our function here, which I will call $h(x) = (x^2-1)^2$, is a composition of two functions. You have the inner function which I will define as a $g(x) = x^2 - 1$ and the outer function which takes this inner function g(x) and raises it two the power of two ($(g(x)^2$). Which is a function of itself that I will define as $f(x)$. The chain rule states $\frac{d}{dx}h(x) = f'(g(x))g'(x)$ In a simplified analogy, you can think about this as multiplying the derivative of the "outside" (keeping the "inside" the same) by the derivative of the inside. Here we go. $\frac{d}{dx} (x^2-1)^2 = 2(x^2-1)^1 \frac{d}{dx}(x^2-1)$ $\Rightarrow 2(x^2-1)^1 2x$ = 4x(x^2-1) Notice in the first step, all I did was use the power rule on the exponent on the outside of the parentheses and kept the stuff inside the parentheses constant. Then I multiplied this result by the derivative of the stuff on the inside of the parentheses. Hope this is clear.
February 24th, 2019, 09:32 AM   #4
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Quote:
 Originally Posted by MikeAndIke Is the answer: (2x^2-2)?
The chain rule is good here. But you can simply expand it:
$\displaystyle (x^2 - 1)^2 = (x^2)^2 + 2 \cdot (x^2) \cdot (-1) + (-1)^2 = x^4 - 2x^2 + 1$.

-Dan

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