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Thread: Help w/ Problems for Test Tomorrow

  1. #1
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    Help w/ Problems for Test Tomorrow

    There are several I need help with.

    Find the derivative

    1. $\displaystyle y=ln(x^2+y^2)$
    2. $\displaystyle f(u)=(2^u+2^{-u})^{10}$
    3. $\displaystyle y=x^x$

    Use the substitution rule
    4. $\displaystyle \int \frac{cos\sqrt{t}}{\sqrt{t}}$
    5. $\displaystyle \int \cos\theta\sin^6\theta d\theta$

    thanks
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  2. #2
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    Quote Originally Posted by c_323_h
    4. $\displaystyle \int \frac{cos\sqrt{t}}{\sqrt{t}}$
    Express as,
    $\displaystyle 2\int \frac{\cos \sqrt{t}}{1} \cdot \frac{1}{2\sqrt{t}} dt$
    Let,
    $\displaystyle u=\sqrt{t}$, thus $\displaystyle \frac{du}{dt}=\frac{1}{2\sqtr{t}}$
    Thus,
    $\displaystyle 2\int \cos u \frac{du}{dt} dt=2\int \cos u du$
    Thus,
    $\displaystyle 2\sin u+C$
    Substitute back,
    $\displaystyle 2\sin \sqrt{t}+C$
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  3. #3
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    Quote Originally Posted by c_323_h
    5. $\displaystyle \int \cos\theta\sin^6\theta d\theta$
    Let,
    $\displaystyle u=\sin \theta$ then, $\displaystyle \frac{du}{d\theta }=\cos \theta$
    Thus,
    $\displaystyle \int u^6 \frac{du}{d\theta} d\theta=\int u^6 du$
    Thus,
    $\displaystyle \frac{1}{7}u^7+C=\frac{1}{7}\sin^7 \theta +C$
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  4. #4
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    Quote Originally Posted by c_323_h
    1. $\displaystyle y=ln(x^2+y^2)$
    IMPLICITLY.
    (Remember to use chain rule now)
    $\displaystyle y'=\frac{(x^2+y^2)'}{x^2+y^2}$
    Thus,
    $\displaystyle y'=\frac{2x+2yy'}{x^2+y^2}$
    Thus,
    $\displaystyle x^2y'+y^2y'=2x+2yy'$
    Thus,
    $\displaystyle x^2y'+y^2y'-2yy'=2x$
    Thus,
    $\displaystyle y'(x^2-2y+y^2)=2x$
    Thus,
    $\displaystyle y'=\frac{x^2-2y+y^2}{2x}$
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  5. #5
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    Quote Originally Posted by c_323_h
    3. $\displaystyle y=x^x$
    Note,
    $\displaystyle x^x=(e^{\ln x})^x=e^{x\ln x}$
    Now just use chain rule,
    $\displaystyle y'=(x\ln x)'e^{x\ln x}$
    Product rule,
    $\displaystyle y'=(1+\ln x)e^{x\ln x}$
    Express back,
    $\displaystyle y'=(1+\ln x)x^x$
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  6. #6
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    Quote Originally Posted by c_323_h
    2. $\displaystyle f(u)=(2^u+2^{-u})^{10}$
    Use chain rule,
    $\displaystyle 10(2^u+2^{-u})'(2^u+2^{-u})^9$
    Thus,
    $\displaystyle 10(\ln 2\cdot 2^u-\ln 2\cdot 2^{-u})(2^u+2^{-u})^9$
    Thus,
    $\displaystyle 10\ln 2(2^u-2^{-u})(2^u+2^{-u})^9$
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  7. #7
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    Thank you very much for the quick response...and for solving all the problems.
    Last edited by c_323_h; Jul 4th 2006 at 03:01 PM.
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  8. #8
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    Hello, c_323_h!

    Find the derivative:

    $\displaystyle 1)\; y \:=\:\ln(x^2+y^2)$

    Differentiate implicitly: .$\displaystyle y' \:=\:\frac{1}{x^2+y^2}(2x + 2xy\cdot y')$

    Then we have: .$\displaystyle (x^2 + y^2)y' \:=\:2x + 2y\cdot y'$

    Rearrange terms: .$\displaystyle x^2\cdot y' + y^2\cdot y' - 2y\cdot y' \:= \:2x$

    Factor: .$\displaystyle (x^2 + y^2 - 2y)y' \:= \:2x$

    Therefore: .$\displaystyle y' \;= \;\frac{2x}{x^2 + y^2 - 2y}$



    $\displaystyle 2)\;f(u)\:=\2^u + 2^{-u})^{10}$

    $\displaystyle f'(u)\;=\;10\left(2^u + 2^{-u}\right)^9\cdot\left(2^u\cdot\ln 2 + 2^{-u}[-1]\cdot\ln 2\right)$

    $\displaystyle f\(u) \;= \;10\cdot\ln 2\left(2^u + 2^{-u}\right)^9\left(2^u - 2^{-u}\right)$


    $\displaystyle 3)\;y\;=\;x^x$

    Take logs: .$\displaystyle \ln(y) \;=\;\ln\left(x^x\right) \;= \;x\cdot\ln x$

    Differentiate implicitly: .$\displaystyle \frac{1}{y}\,y' \;= \;x\cdot\frac{1}{x} + 1\cdot\ln x$

    We have: .$\displaystyle \frac{y'}{y}\;=\;1 + \ln x\quad\Rightarrow\quad y' \;= \;y(1 + \ln x)$

    Since $\displaystyle y = x^x$, we have: .$\displaystyle y' \;= \;x^x(1 + \ln x)$



    Use the substitution rule:

    $\displaystyle 4)\;\int \frac{\cos\sqrt{t}}{\sqrt{t}}\,dt$

    We have: .$\displaystyle \int t^{-\frac{1}{2}}\cdot\cos\left(t^{\frac{1}{2}}\right) \,dt $

    Let: $\displaystyle u = t^{\frac{1}{2}}\quad\Rightarrow\quad du = \frac{1}{2}t^{-\frac{1}{2}}dt\quad\Rightarrow\quad dt = 2t^{\frac{1}{2}}du$

    Substitute: .$\displaystyle \int t^{-\frac{1}{2}}\cdot\cos(u)\cdot\left(2t^{\frac{1}{2} }du\right) \;= \;2\int\cos u\,du$

    You can finish it now . . .



    $\displaystyle 5)\;\int \cos\theta\sin^6\theta d\theta$

    We have: .$\displaystyle \int(\sin\theta)^6(\cos\theta\,d\theta)$

    Let: $\displaystyle u = \sin\theta\quad\Rightarrow\quad du = \cos\theta\, d\theta\quad\Rightarrow\quad d\theta$$\displaystyle = \frac{du}{\cos\theta}$

    Substitute: .$\displaystyle \int u^6\,\cos\theta\,\frac{du}{\cos\theta} \;= \;\int u^6\,du$ . . . got it?

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  9. #9
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    thanks you. as you can tell i am struggling with this class. is anyone willing to help me with these?

    Differentiate:

    1. $\displaystyle y=cos^{-1}(e^{2x})$
    2. $\displaystyle h(t)=cot^{-1}(t)+cot{-1}(\frac{1}{t})$

    Evaluate the integral:

    1. $\displaystyle \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}$
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  10. #10
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    Quote Originally Posted by c_323_h

    1. $\displaystyle \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}$
    Manipulate,
    $\displaystyle \int \frac{6}{t-t^2} dt$
    As, (since $\displaystyle t\geq 0$) (I factored it)

    $\displaystyle 12 \int \cdot \frac{1}{\sqrt{1-t}}\cdot \frac{1}{2\sqrt{t}} dt$

    Let, $\displaystyle u=\sqrt{t}$ then, $\displaystyle \frac{du}{dt} =\frac{1}{2\sqrt{t}}$.
    Thus, far we have,
    $\displaystyle \int \frac{1}{\sqrt{t-1}} \frac{du}{dt} dt$

    Since, $\displaystyle u=\sqrt{t}$ then, $\displaystyle u^2=t$ thus, $\displaystyle u^2-1=t-1$ thus, $\displaystyle \sqrt{u^2-1}=\sqrt{t-1}$ Thus, we have,

    $\displaystyle \int \frac{1}{\sqrt{u^2-1}} \frac{du}{dt}dt=\int \frac{1}{\sqrt{u^2-1}} du$

    This is a inverse hyperbolic function which you should recognize thus,
    $\displaystyle \cosh^{-1} u+C$
    Substitute back,
    $\displaystyle \cosh^{-1} \sqrt{t}+C$
    Thus,
    $\displaystyle \left \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}=\cosh^{-1} \sqrt{t} \right|^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}$
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  11. #11
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    Quote Originally Posted by ThePerfectHacker
    $\displaystyle x^2y'+y^2y'=2x+2yy'$
    Could you tell me where you got the $\displaystyle y'$ from on the right side of the equation?
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  12. #12
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    Hello again, c_323_h!

    Differentiate:

    $\displaystyle 1)\;y\:=\:\cos^{-1}(e^{2x})$
    $\displaystyle y' \;= \;\frac{-1}{\sqrt{1 - \left(e^{2x}\right)^2}}\cdot e^{2x}\cdot2 \;= \;\frac{-2e^{2x}}{\sqrt{1 - e^{4x}}}$


    $\displaystyle 2)\;h(t)\:=\:\cot^{-1}(t) + \cot^{-1}\left(\frac{1}{t}\right)$
    $\displaystyle h'(t)\;=\;\frac{-1}{1 + t^2} + \frac{-1}{1 + \left(\frac{1}{t}\right)^2}\cdot(-t^{-2}) \;= \;\frac{-1}{1 + t^2} + \frac{1}{\left(1 + \frac{1}{t^2}\right)t^2} $

    . . . . . $\displaystyle =\;-\frac{1}{1 + t^2} + \frac{1}{1 + t^2} \;= \;0$ **


    Evaluate the integral: .$\displaystyle 1)\;\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}$

    In the denominator, complete the square:
    $\displaystyle t - t^2\;=\;-(t^2 - t) \;= \;-\left(t^2 - t + \frac{1}{4} - \frac{1}{4}\right) \;=$ $\displaystyle -\left(\left[t - \frac{1}{2}\right]^2 - \frac{1}{4}\right) \;= \;\frac{1}{4} - \left(x - \frac{1}{2}\right)^2$

    The integral becomes: .$\displaystyle 6\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} \frac{1}{\sqrt{\frac{1}{4} - (t - \frac{1}{2})^2}}\,dt$

    . . Let $\displaystyle t - \frac{1}{2}\:=\:\frac{1}{2}\sin\theta\quad \Rightarrow\quad dt\,=\,\frac{1}{2}\cos\theta\,d\theta$

    . . Note that: .$\displaystyle \sqrt{\frac{1}{4} - \left(t - \frac{1}{2}\right)^2} \;= \;\sqrt{\frac{1}{4} - \frac{1}{4}\sin^2\theta} \;=$ $\displaystyle \sqrt{\frac{1}{4}\left(1 - \sin^2\theta)}$

    . . . . . . . . . . $\displaystyle =\;\sqrt{\frac{1}{4}\cos^2\theta}\;=\;\frac{1}{2} \cos\theta$

    Substitute: .$\displaystyle 6\int\frac{1}{\frac{1}{2}\cos\theta}\cdot\frac{1}{ 2} \cos\theta\,d\theta \;= \;6\int d\theta\;= \;6\theta$

    Back-substitute
    We have: .$\displaystyle \frac{1}{2}\sin\theta \,= \,t - \frac{1}{2}\quad\Rightarrow\quad \sin\theta\,=\,2t - 1\quad\Rightarrow\quad\theta \,= \,\sin^{-1}(2t-1)$

    So we have: .$\displaystyle 6\sin^{-1}(2t - 1)\,\bigg]^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} \;= $ $\displaystyle \;6\sin^{-1}\left[2\left(\frac{\sqrt{3}}{2}\right) - 1\right] - 6\sin^{-1}\left[2\left(\frac{1}{2}\right) - 1\right]$

    . . . $\displaystyle = \;6\sin^{-1}(\sqrt{3} - 1) - 6\sin^{-1}(0) \;= \;6\sin^{-1}(\sqrt{3} - 1) \;\approx\; 4.928$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    In #2, we have: .$\displaystyle h(t) \:=\:\cot^{-1}(t) + \cot^{-1}\left(\frac{1}{t}\right)$

    Let: $\displaystyle \cot^{-1}(t) = \alpha$ . . . and: $\displaystyle \cot^{-1}(\frac{1}{t}) = \beta$

    Then $\displaystyle \cot\alpha = t$ . . . and $\displaystyle \cot\beta = \frac{1}{t}$

    Since $\displaystyle \cot\alpha = \frac{t}{1} = \frac{adj}{opp}$, $\displaystyle \alpha$ is in this right triangle:
    Code:
                  *
                /β|
              /   |1
            /α    |
          * - - - *
              t

    Since $\displaystyle \cot\beta = \frac{1}{t},\;\beta$ is the other acute angle.

    That is: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2}$

    Hence, the function is: .$\displaystyle h(t)\;=\;\cot^{-1}(t) + \cot^{-1}\left(\frac{1}{t}\right) \;= \;\alpha + \beta \;= \;\frac{\pi}{2}$

    No wonder the derivative is 0 . . .

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