Homework 4 - Solution

Date: MATH 1100-2 - Spring 2002

1. 9.7.5. Using simple properties of addition and multiplication by constants, as well as the derivative of $x^n$, we have:

   $$\begin{split}g'(x) &= (5x^3+\frac{4}{x})' = (5x^3)' +(4\cdot ... ... x^2 + 4 \cdot(-1) x^{-2} = 15 x^2 - \frac{4}{x^2}. \end{split}$$

   
   

2. 9.7.7. For $y=(x^2-2)(x+4)$, using the product rule we have:

   $$\begin{split}y' &= ((x^2-2)(x+4))' = (x^2-2)'(x+4) + (x^2-2)(... ...)') = (2x-0)(x+4)+(x^2-2)(1+0) \\ &= 2x(x+4)+x^2-2 \end{split}$$

   
   

3. 9.7.17. For $y=\frac{(x^2-4)^3}{x^2+1}$, we use the quotient rule first and then the chain rule:

   $$\begin{split}y' &= (\frac{(x^2-4)^3}{x^2+1})' = \frac{((x^2-4... ...x^2-4)^2\cdot 2x (x^2+1) - (x^2-4)^3 2x}{(x^2+1)^2} \end{split}$$

   
   

4. 9.7.33.

   1. $$\begin{split}\frac{dF_1}{dx} &= \frac{d}{dx} \overbrace{\frac... ...(x^4+1)}{dx} = 3u^4\cdot 4x^3 \\ &= 12x^3(x^4+1)^4 \end{split}$$

      
      

   2. $$\begin{split}\frac{dF_2}{dx} &= \frac{d}{dx} \underbrace{\fra... ...{dx} \\ &= -3u^{-6}\cdot 4x^3 = -12x^3(x^4+1)^{-6} \end{split}$$

      
      

   3. $$\begin{split}\frac{dF_3}{dx} &= \frac{d}{dx} \overbrace{\frac... ...}{dx} = u^4\cdot 3\cdot 4x^3 \\ &= 12x^3(3x^4+1)^4 \end{split}$$

      
      

   4. $$\begin{split}\frac{dF_4}{dx} &= \frac{d}{dx} \underbrace{\fra... ... &= -3u^{-6}\cdot 5\cdot 4x^3 = -60x^3(5x^4+1)^{-6} \end{split}$$

      
      

5. 9.8.1. For $f(x)= 4x^3-15x^2+3x+2$, we first compute

   $$\begin{split}f'(x) &= (4x^3-15x^2+3x+2)' = (4x^3)'-(15x^2)'+(... ...+0= 4\cdot 3x^2-15\cdot 2x +3\cdot 1 = 12x^2-30x+3. \end{split}$$

   
   
   Then:

   $$\begin{split}f''(x) &= (f'(x))' = (12x^2-30x+3)' = (12x^2)'-(... ...2)'-30(x)'+ 0 \\ &= 12\cdot 2 x-30\cdot 1 = 24x-30 \end{split}$$

   
   

6. 9.8.7. For $y=x^3-\sqrt{x} = x^3-x^{\frac{1}{2}}$ we first find

   $$y'= (x^3-x^{\frac{1}{2}})' = (x^3)'-(x^{\frac{1}{2}})' =3x^2 - \frac{1}{2} x^{-\frac{1}{2}}.$$

   
   
   Then

   $$\begin{split}y'' &= (y')' = (3x^2 - \frac{1}{2} x^{-\frac{1}{... ...x^{-\frac{3}{2}} = 6x + \frac{1}{4}x^{-\frac{3}{2}} \end{split}$$

   
   

7. 9.8.27. We have that $f^{(4)}(x) = x(x+1)^{-1}= \frac{x}{x+1}$. Then:

   $$\begin{split}f^{(5)}(x) &= (f^{(4)}(x))' = (\frac{x}{x+1})' =... ... 1}{(x+1)^2} \\ &= \frac{1}{(x+1)^2} = (x+1)^{-2}. \end{split}$$

   
   
   And then:

   $$\begin{split}f^{(6)}(x) &= (f^{(5)}(x))' = ((x+1)^{-2})' = \f... ...c{d(x+1)}{dx} = -2(x+1)^{-3}\cdot 1 = -2(x+1)^{-3}. \end{split}$$