Why is Taking a Derivative of Quantities to a Negative Exponent an Application of the Chain Rule, Not the Power Rule

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According to my calculus professor and MIT open coursework, taking the derivative of (x^2+4)^-1 is an application of the chain rule, not the power rule. The answer to the question is -(x^2+4)^-2, which makes sense to me, but I just don't understand why this is considered an application of the chain rule rather than the power rule, since the power rule says that d/dx(x^n) = nx^(n-1).

Here is a link to the MIT coursework I am talking about: https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-11-chain-rule/MIT18_01SCF10_ex11sol.pdf

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The Power Rule was applied, but so too was the chain rule, and also a few unmentioned others.

The power rule may only be applied when the derivation 'numerator' is a power of the 'denominator' , so first you must use the Chain Rule to obtain such.

Thus taking everything one elementary step at a time:

$$\begin{align}\dfrac{\mathrm d~(x^2+4)^{-1}}{\mathrm d~x}&=\dfrac{\mathrm d ~(x^2+4)^{-1}}{\mathrm d~(x^2+4)}\cdot\dfrac{\mathrm d~(x^2+4)}{\mathrm d~x}&&\text{Chain Rule}\\[2ex]&=-(x^2+4)^{-2}\cdot\dfrac{\mathrm d~(x^2+4)}{\mathrm d~x}&&\text{Power Rule}\\[2ex]&=-(x^2+4)^{-2}\cdot\left(\dfrac{\mathrm d~x^2}{\mathrm d~x}+\dfrac{\mathrm d~4}{\mathrm d~x}\right)&&\text{Additive Rule}\\[2ex]&=-(x^2+4)^{-2}\cdot\left(2x+\dfrac{\mathrm d~4}{\mathrm d~x}\right)&&\text{Power Rule }\\[2ex]&=-(x^2+4)^{-2}\cdot\left(2x+0\right)&&\text{Derivation of a Constant}\\[2ex]&= -2x(x^2+4)^{-2} &&\text{Algebraic Rearrangement}\end{align}$$

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The derivative is $-2x(x^2+4)^{-2}$. Thus the chain rule. You missed $2x$.