## Calculus: Early Transcendentals 8th Edition

(a) Using Quotient Rule normally $$\frac{d}{dx}(\frac{a}{b})=\frac{a'b-ab'}{b^2}$$ with $a=1$ and $b=g(x)$ in this case. Then simplify to get the results wanted. (b) $$y'=\frac{3t^2+4t}{(t^3+2t^2-1)^2}$$ (c) Rewrite $x^{-n}$ into $\frac{1}{x^n}$ to be able to apply the Reciprocal Rule. Then simplify.
(a) According to Quotient Rule, we have $$\frac{d}{dx}\Bigg[\frac{1}{g(x)}\Bigg]=\frac{(1)'g(x)-1g'(x)}{[g(x)]^2}$$ $$\frac{d}{dx}\Bigg[\frac{1}{g(x)}\Bigg]=\frac{0g(x)-g'(x)}{[g(x)]^2}$$ $$\frac{d}{dx}\Bigg[\frac{1}{g(x)}\Bigg]=-\frac{g'(x)}{[g(x)]^2}$$ The statement has been proved. b) $y= \frac{1}{t^3+2t^2-1}$ Using Reciprocal Rule to differentiate the function: $$y'=-\frac{(t^3+2t^2-1)'}{(t^3+2t^2-1)^2}$$ $$y'=\frac{3t^2+4t}{(t^3+2t^2-1)^2}$$ (c) $$\frac{d}{dx}(x^{-n})=\frac{d}{dx}(\frac{1}{x^n})$$ Applying Reciprocal Rule, we have $$\frac{d}{dx}(x^{-n})=-\frac{(x^n)'}{(x^n)^2}$$ $$\frac{d}{dx}(x^{-n})=-\frac{nx^{n-1}}{x^{2n}}$$ $$\frac{d}{dx}(x^{-n})=-nx^{n-1-2n}$$ $$\frac{d}{dx}(x^{-n})=-nx^{-n-1}$$