Homework 3 - Solution


Date: MATH 1100-2 - Spring 2002

  1. 9.4.1. $ y=4 \ensuremath{\Rightarrow}\xspace y' = (4)' = 0$. (The derivative of a constant function is 0.)

  2. 9.4.3. $ y=x \ensuremath{\Rightarrow}\xspace y' = (x)' = 1$. (This is the derivative of a power.)

  3. 9.4.5. $ f(x)=2x^3-x^5$, so

    $\displaystyle f'(x) = (2x^3-x^5)' = (2x^3)'-(x^5)' = 2(x^3)'-5x^4 = 2\cdot 3x^2-5x^4 = 6x^2-5x^4.$    

  4. 9.4.7. $ y=6x^4-5x^2+x-2$, so:

    \begin{displaymath}\begin{split}y' &= (6x^4-5x^2+x-2)' = (6x^4)'-(5x^2)'+(x)'-(2... ...-0 \\ &= 6\cdot 4x^3-5\cdot2 x +1 = 24x^3-10 x +1. \end{split}\end{displaymath}    

  5. 9.4.15. $ y=x^{-5}+x^{-8}-3$, so:

    $\displaystyle y' = (x^{-5}+x^{-8}-3)' = (x^{-5})'+(x^{-8})'-(3)' = -5 x^{-5-1} + (-8) x^{-8-1} -0 = -5x^{-6} - 8 x^{-9}.$    

  6. 9.4.17. $ y=3x^{11/3} -2x^{7/4}-x^{1/2}+8$, so:

    \begin{displaymath}\begin{split}y' &= (3x^{11/3} -2x^{7/4}-x^{1/2}+8)' = (3x^{11... ...x^{8/3} - \frac{7}{2} x^{3/4}- \frac{1}{2}x^{-1/2}. \end{split}\end{displaymath}    

  7. 9.4.23. The equation of the tangent line to the graph of $ f(x)$ at $ x_0$ is $ y=f'(x_0)(x-x_0) + f(x_0)$, or in terms of $ y(x)$, $ y= y'(x_0) (x-x_0) + y(x_0)$.

    We start by computing $ y'(x)$.

    $\displaystyle y'= (x^3-3x^2+5)' = (x^3)'-(3x^2)'+(5)' = 3x^2- 3(x^2)'+0 = 3x^2-6x.$    

    Then, $ y'(1)= 3\cdot 1^2-6\cdot 1 = -3$, and $ y(1) = 1^3-3\cdot 1^2 +5 = 3$.

    All together, the equation of the tangent line at $ x_0 = 1$ is $ y = -3(x-1) + 3 = -3x+6$.

  8. 9.5.1. $ y=(x+3)(x^2-2x)$. We use the product rule and obtain:

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

  9. 9.5.5. $ f(x)=(x^{12}+3x^4+4)(4x^3-1)$. We use the product rule and obtain:

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

  10. 9.5.9. $ y=(x^2+x+1)(\sqrt[3]{x}-2\sqrt{x}+5) = (x^2+x+1)(x^{1/3}-2x^{1/2}+5)$. We use the product rule and obtain:

    \begin{displaymath}\begin{split}y' &= ((x^2+x+1)(x^{1/3}-2x^{1/2}+5))' \\ &= (x... ... (\frac{1}{3} x^{-2/3}-2 \cdot \frac{1}{2}x^{-1/2}) \end{split}\end{displaymath}    

  11. 9.5.13. $ y = \frac{x}{x^2-1}$. Using the quotient rule we obtain:

    \begin{displaymath}\begin{split}y' &= (\frac{x}{x^2-1})' = \frac{(x)'(x^2-1) - x... ...2x-0)}{(x^2-1)^2} = \frac{x^2-1 - 2x^2}{(x^2-1)^2}. \end{split}\end{displaymath}    

  12. 9.5.25.
    1. The slope of the tangent line at $ x=2$ is the derivative $ y'(2)$. We start by computing the derivative $ y'(x)$:

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

      We now plug in $ x=2$ and get the slope at the point is $ y'(2) = \frac{3\cdot 2^2+6\cdot2+1}{(2+3)^2} = \frac{25}{25} = 1$.
    2. The instantaneous rate of change of $ y$ at $ x=2$ is precisely the derivative $ y'(2) = 1$, as we saw above.

  13. 9.6.5. We think of $ f(x)=\frac{1}{(x^2+2)^3}$ as $ f(x) = g(h(x))$, where $ g(x) = \frac{1}{x^3} = x^{-3}$ and $ h(x) = x^2+2$. In order to find $ f'(x)$ I have to use the chain rule ( $ f'(x)=g'(h(x))\cdot h'(x)$), so that I need to know $ g'$ and $ h'$ first.
    $\displaystyle g'(x)$ $\displaystyle =$ $\displaystyle (x^{-3})' = -3x^{-4}$  
    $\displaystyle h'(x)$ $\displaystyle =$ $\displaystyle (x^2+2)' = (x^2)'+(2)' = 2x+0 = 2x$  

    Now we use the chain rule:

    $\displaystyle f'(x) = g'(h(x))\cdot h'(x) = -3(h(x))^{-4}\cdot 2x = -3(x^2+2)^{-4}\cdot 2x = -\frac{6x}{(x^2+2)^4}.$    

  14. 9.6.15. We think of $ s(x)=4\sqrt{3x-x^2}$ as $ s(x)=g(h(x))$, where $ g(x) = 4\sqrt{x} = 4x^{1/2}$ and $ h(x) = 3x-x^2$. In order to find $ s'(x)$ I have to use the chain rule ( $ s'(x)=g'(h(x))\cdot h'(x)$), so that I need to know $ g'$ and $ h'$ first.
    $\displaystyle g'(x)$ $\displaystyle =$ $\displaystyle (4x^{1/2})' = 4(x^{1/2})' = 4\cdot \frac{1}{2} x^{-1/2} = \frac{2}{x^{1/2}}$  
    $\displaystyle h'(x)$ $\displaystyle =$ $\displaystyle (3x-x^2)' = (3x)'-(x^2)'= 3(x)' - 2x = 3-2x$  

    Now we use the chain rule:

    $\displaystyle s'(x) = g'(h(x))\cdot h'(x) = \frac{2}{(h(x))^{1/2}}\cdot (3-2x) = \frac{2(3-2x)}{\sqrt{3x-x^2}}.$    

  15. 9.6.27. The equation of the tangent line to the graph of $ f(x)$ at the point $ x_0$ is $ y=f'(x_0)(x-x_0) + f(x_0)$. Thus we need to know the slope of the line, that is, the derivative of the function at $ x_0$, $ f'(x_0) = y'(x_0)$. In our case, $ x_0 = 3$.

    We start by finding the derivative $ y'(x)$. For that, we think of $ y$ as $ y(x)=g(h(x))$, for $ g(x)=\sqrt{x} = x^{1/2}$ and $ h(x) = 3x^2-2$. In order to use the chain rule, we compute

    $\displaystyle g'(x)$ $\displaystyle =$ $\displaystyle (x^{1/2})' = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$  
    $\displaystyle h'(x)$ $\displaystyle =$ $\displaystyle (3x^2-2)' = (3x^2)'-(2)' = 3(x^2)' - 0 =3\cdot 2 x =6x$  

    Next we use the chain rule:

    $\displaystyle y'(x) = g'(h(x))\cdot h'(x) = \frac{1}{2\sqrt{h(x)}}\cdot 6x = \frac{3x}{\sqrt{3x^2-2}}$    

    Then, the slope of the tangent line at the point $ x=3$ is $ y'(3) = \frac{3\cdot3 }{\sqrt{3\cdot 3^2-2}} = \frac{9}{5}$. Last, $ y(3)= \sqrt{3\cdot3^2-2} = 5$, so that the equation of the tangent line at $ x=3$ is

    $\displaystyle y=y'(3)(x-3)+y(3) = \frac{9}{5}(x-3)+5$    


Javier Fernandez
2002-01-30