## Calculus: Early Transcendentals 8th Edition

(a) $$y'=2xf(x)+x^2f'(x)$$ (b) $$y'=\frac{xf'(x)-2f(x)}{x^3}$$ (c) $$y'=\frac{2xf(x)-x^2f'(x)}{[f(x)]^2}$$ (d) $$y'=-\frac{1}{2\sqrt {x^3}}+\frac{f(x)}{2\sqrt x}+\sqrt xf'(x)$$
Since $f$ is a differentiable function, $f'(x)$ exists. (a) $y=x^2f(x)$ The derivative of $y$ is $$y'=(x^2)'f(x)+x^2f'(x)$$ $$y'=2xf(x)+x^2f'(x)$$ (b) $y=\frac{f(x)}{x^2}$ The derivative of $y$ is $$y'=\frac{f'(x)x^2-f(x)(x^2)'}{(x^2)^2}$$ $$y'=\frac{x^2f'(x)-2xf(x)}{x^4}$$ $$y'=\frac{xf'(x)-2f(x)}{x^3}$$ (c) $y=\frac{x^2}{f(x)}$ The derivative of $y$ is $$y'=\frac{(x^2)'f(x)-(x^2)f'(x)}{[f(x)]^2}$$ $$y'=\frac{2xf(x)-x^2f'(x)}{[f(x)]^2}$$ (d) $$y=\frac{1+xf(x)}{\sqrt x}$$ $$y=\frac{1}{\sqrt x}+\frac{xf(x)}{\sqrt x}$$ $$y=\frac{1}{x^{1/2}}+\sqrt xf(x)$$ $$y=x^{-1/2}+x^{1/2}f(x)$$ The derivative of $y$ is $$y'=(x^{-1/2})'+(x^{1/2}f(x))'$$ $$y'=-\frac{1}{2}x^{-3/2}+(x^{1/2})'f(x)+x^{1/2}f'(x)$$ $$y'=-\frac{1}{2}x^{-3/2}+\frac{1}{2}x^{-1/2}f(x)+x^{1/2}f'(x)$$ $$y'=-\frac{1}{2\sqrt {x^3}}+\frac{f(x)}{2\sqrt x}+\sqrt xf'(x)$$