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(a) If $ g $ is differentiable. the Reciprocal Rule says that

$ \frac {d}{dx} \left[ \frac {1}{g(x)} \right] = - \frac {g'(x)}{[g(x)]^2} $

Use the Quotient Rule to prove the Reciprocal Rule.

(b) Use the Reciprocal Rule lo differentiate the function in Exercise 16.

(c) Use the Reciprocal Rule to verify that the Power Rule is valid for negative integers, that is,

$ \frac {d}{dx} (x^{-n}) = - nx^{-n-1} $

for all positive integers $ n. $

(a) $-\frac{g^{\prime}(x)}{\left[g^{(x)}\right]^{2}}$

(b) $y=\frac{1}{g(x)}$

(c) $-n x^{-n-1}$

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he had seen Marie Claire. So name right here. So there g is different shovel. Then we could use the question rule. So we have de over de luxe won over G of X, but just equal to G of X. Do you? Over DX of one minus D over D'backs of G of X, all over G of X square, which is equal to negative. The derivative of G of X over G of X square for a part B we're going to have why it is equal to one over X cute plus two X square minus one. We're gonna make this equal the denominator B g of X. So the derivative is going to be equal to three X square plus four of X. So we get why is equal to one over g of X. So we're gonna referring to the recipe. Oh, girl, we just plug it in when this becomes equal to negative Three jacks square plus four x over X cubed plus two X square, minus one square for part C. We see we got this equation from part, eh? For a different for a differential function G of X. So he is G of X is equal to X to the end and is a positive integer. So we get D over D X worn over X to the end, this sequel to negative extent and the derivative over next to that one square. This part is the power all for when n is a positive integer. So we get X to the negative end, which is equal to negative, then turns X to the N minus one over X to the two and which becomes equal to negative and cracks. It's the negative and minus one.