91Ó°ÊÓ

Consider the function \(k\) given by \(k(x)=2|x+5|\) a) For what \(x\) -value(s) is this function not differentiable? b) Evaluate \(k^{\prime}(-10), k^{\prime}(-7), k^{\prime}(-2),\) and \(k^{\prime}(0) .\) Is there a shortcut you can use to find these slopes?

Find \(y^{\prime}\). $$\text { If } f(x)=\sqrt{x}, \text { find } f^{\prime}(4)$$

Find \(\frac{d y}{d x}\) for each pair of functions.\( $$y=u^{3}-7 u^{2}\) and $u=x^{2}+3$$

Let \(f(x)=\frac{x^{2}+4 x+3}{x+1} .\) A student recognizes that this function can be simplified as follows: $$f(x)=\frac{x^{2}+4 x+3}{x+1}=\frac{(x+1)(x+3)}{x+1}=x+3$$ since \(y=x+3\) is a line with slope \(1,\) the student makes the lollowing conclusions: \(f^{\prime}(-2)=1, f^{\prime}(-1)=1\) \(f^{\prime}(0)=1, \int^{\prime}(1)=1 .\) Where did the student make an error?

Find the simplified difference quotient for each function listed. $$f(x)=\frac{1}{1-x}$$

Find \(y^{\prime}\). $$\text { If } y=\frac{4}{x^{2}}, \text { find }\left.\frac{d y}{d x}\right|_{x=-2}$$

Is the function given by \(g(x)=\left\\{\begin{array}{ll}\frac{1}{3} x+4 & \text { for } x \leq 3, \\ 2 x-1 & \text { for } x>3.\end{array}\right.\) continuous at \(x=3 ?\) Why or why not?

Find \(\frac{d y}{d x}\) for each pair of functions.\( $$y=\sqrt[3]{2 u+5}\) and $u=x^{2}-x$$

Find \(y^{\prime}\). $$\text { If } y=x+\frac{2}{x^{3}}, \text { find }\left.\frac{d y}{d x}\right|_{x=1}$$

Find \(\frac{d y}{d x}\) for each pair of functions.\( $$y=\sqrt{7-3 u}\) and $u=x^{2}-9$$