Homework 5 - Solution

Date: MATH 1100-2 - Spring 2002

1. 11.1.1. For $f(x)=4\ln(x)$:

   $$f'(x) = (4\ln(x))' = 4(\ln(x))' = 4\cdot\frac{1}{x} = \frac{4}{x}.$$

   
   

2. 11.1.3. For $y=\ln(8x)$:

   $$\frac{dy}{dx} = \frac{d}{dx}\underbrace{\ln(8x)}_{u=8x} = \frac{d... ...ln(u)}{du}\cdot\frac{du}{dx} = \frac{1}{u} \cdot 8 = \frac{8}{8x} = \frac{1}{x}$$

   
   

3. 11.1.5. For $y=\ln(x^4)$:

   $$\frac{dy}{dx} = \frac{d}{dx}\underbrace{\ln(x^4)}_{u=x^4} = \frac... ...dx} = \frac{1}{u}\cdot \frac{d x^4}{dx} = \frac{1}{x^4}\cdot 4x^3 = \frac{4}{x}$$

   
   
   A different alternative would be using that $\ln(x^4) = 4 \ln(x)$ and then look at exercise 11.1.1.

4. 11.1.11. $p = \ln(q^2+1)$, so:

   $$\frac{dp}{dq} = \frac{d}{dq} \underbrace{\ln(q^2+1)}_{u=q^2+1} = ... ...c{1}{u}\cdot \frac{d(q^2+1)}{dq} = \frac{1}{q^2+1}\cdot 2q = \frac{2q}{ q^2+1}.$$

   
   

5. 11.1.21. $p = \ln(\frac{q^2-1}{q})$. You can do it in two ways: using properties of the logarithm: $p = \ln(q^2-1) - \ln(q)$, so that

   $$\begin{split}\frac{dp}{dq} &= \frac{d}{dq} (\ln(q^2-1) - \ln(... ...(2q) - \frac{1}{q} = \frac{2q}{q^2-1} - \frac{1}{q} \end{split}$$

   
   
   Also,

   $$\begin{split}\frac{dp}{dq} &= \frac{d}{dq} \underbrace{\ln(\f... ...} \cdot \frac{q^2+1}{q^2} = \frac{q^2+1}{q(q^2-1)}. \end{split}$$

   
   
   A little algebraic massage shows that the two results agree.

6. 11.1.33. For $y=(\ln((x^4+3)))^2$:

   $$\begin{split}\frac{dy}{dx} &= \frac{d}{dx}\underbrace{(\ln((x... ...4+3} \cdot 4 x^3 = 8\frac{ \ln((x^4+3))x^3}{x^4+3}. \end{split}$$

   
   

7. 11.2.1. For $y=5e^x-x$,

   $$y' = (5e^x-x)' = (5e^x)'-(x)' = 5(e^x)'-1= 5e^x-1.$$

   
   

8. 11.2.7. For $y=6e^{3x^2}$ we have:

   $$\frac{dy}{dx} = \frac{d}{dx}\underbrace{6e^{3x^2}}_{u=3x^2} = \fr... ...u}{dx} = 6\frac{de^u}{du}\cdot\frac{d(3x^2)}{dx} = 6e^u\cdot 6x = 36x e^{3x^2}.$$

   
   

9. 11.2.13. Given $y=e^{-\frac{1}{x}}$ we compute

   $$\frac{dy}{dx} = \frac{d}{dx} \underbrace{e^{-\frac{1}{x}}}_{u=-\f... ...}\cdot \frac{du}{dx} = e^u\cdot (\frac{1}{x^2}) = \frac{e^{-\frac{1}{x}}}{x^2}.$$

   
   

10. 11.2.21. For $y=\ln(e^{4x}+2)$:

    $$\begin{split}\frac{dy}{dx} &= \frac{d}{dx} \underbrace{\ln(e^... ... e^v\cdot 4 = \frac{1}{e^{4x}+2}\cdot e^{4x}\cdot 4 \end{split}$$

    
    

11. 11.2.29. For $y=6^x$, we remember that $6^x = (e^{\ln(6)})^x = e^{\ln(6)\cdot x}$. Then:

    $$\frac{dy}{dx} = \frac{d}{dx} \underbrace{e^{\ln(6)\cdot x}}_{u=\l... ...frac{d(\ln(6)x)}{dx} = e^{\ln(6)x}\cdot \ln(6) \frac{dx}{dx} = \ln(6)\cdot 6^x.$$

    
    

12. 11.3.3. We start from the equation $xy^2=8$ and take derivatives on both sides. On the right hand side we have $\frac{d8}{dx}=0$. On the left hand side:

    $$\frac{d}{dx} (xy^2) = \frac{dx}{dx} \cdot y^2 + x \frac{dy^2}{dx}... ...\cdot\frac{dy}{dx} = y^2 +x \cdot 2y\cdot \frac{dy}{dx} = y^2+2xy\frac{dy}{dx}.$$

    
    
    Then, equating the (derivatives of the) left and right hand side:

    $$0 = y^2+2xy\frac{dy}{dx} \ensuremath{\Rightarrow}\xspace \frac{dy}{dx} = -\frac{y^2}{2xy} = -\frac{y}{2x}.$$

    
    

    Now we evaluate the derivative at $x=2$ and $y=2$:

    $$\frac{dy}{dx} = -\frac{2}{2\cdot 2} = -\frac{1}{2}.$$

    
    

13. 11.3.17. From the equation $3x^5-5y^3= 5x^2+3y^5$, we compute the derivative of the right hand side:

    $$\frac{d}{dx}(5x^2+3y^5) = \frac{d 5x^2}{dx} + \frac{d 3y^5}{dx} =... ... 3 \frac{dy^5}{dy}\cdot \frac{dy}{dx} = 10x + 3\cdot 5 y^4 \cdot \frac{dy}{dx},$$

    
    
    and the left hand side:

    $$\frac{d}{dx}(3x^5-5y^3) = \frac{d 3x^5}{dx} - \frac{d5y^3}{dx} = ... ...frac{d y^3}{dy}\cdot \frac{dy}{dx} = 15 x^4 - 3\cdot 5 y^2 \cdot \frac{dy}{dx}.$$

    
    
    Now we equate the two derivatives and solve for $\frac{dy}{dx}$:
    

    $$15 x^4 - 15y^2 \cdot \frac{dy}{dx} = 10x + 15 y^4 \cdot \frac{dy}{dx}$$

    $$15 x^4-10x = 15y^4 \frac{dy}{dx}+15y^2 \frac{dy}{dx}$$

    $$15 x^4-10x = (15y^4+15y^2) \frac{dy}{dx}$$

    $$\frac{15 x^4-10x}{15y^4+15y^2} = \frac{dy}{dx}.$$

    
    
    Thus:

    $$\frac{dy}{dx} = \frac{15 x^4-10x}{15y^4+15y^2} = \frac{3 x^4-2x}{3y^4+3y^2}.$$