Help w/ Problems for Test Tomorrow

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• July 4th 2006, 01:40 PM
c_323_h
Help w/ Problems for Test Tomorrow
There are several I need help with.

Find the derivative

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

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

thanks
• July 4th 2006, 02:04 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
4. $\int \frac{cos\sqrt{t}}{\sqrt{t}}$

Express as,
$2\int \frac{\cos \sqrt{t}}{1} \cdot \frac{1}{2\sqrt{t}} dt$
Let,
$u=\sqrt{t}$, thus $\frac{du}{dt}=\frac{1}{2\sqtr{t}}$
Thus,
$2\int \cos u \frac{du}{dt} dt=2\int \cos u du$
Thus,
$2\sin u+C$
Substitute back,
$2\sin \sqrt{t}+C$
• July 4th 2006, 02:07 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
5. $\int \cos\theta\sin^6\theta d\theta$

Let,
$u=\sin \theta$ then, $\frac{du}{d\theta }=\cos \theta$
Thus,
$\int u^6 \frac{du}{d\theta} d\theta=\int u^6 du$
Thus,
$\frac{1}{7}u^7+C=\frac{1}{7}\sin^7 \theta +C$
• July 4th 2006, 02:17 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
1. $y=ln(x^2+y^2)$

IMPLICITLY.
(Remember to use chain rule now)
$y'=\frac{(x^2+y^2)'}{x^2+y^2}$
Thus,
$y'=\frac{2x+2yy'}{x^2+y^2}$
Thus,
$x^2y'+y^2y'=2x+2yy'$
Thus,
$x^2y'+y^2y'-2yy'=2x$
Thus,
$y'(x^2-2y+y^2)=2x$
Thus,
$y'=\frac{x^2-2y+y^2}{2x}$
• July 4th 2006, 02:19 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
3. $y=x^x$

Note,
$x^x=(e^{\ln x})^x=e^{x\ln x}$
Now just use chain rule,
$y'=(x\ln x)'e^{x\ln x}$
Product rule,
$y'=(1+\ln x)e^{x\ln x}$
Express back,
$y'=(1+\ln x)x^x$
• July 4th 2006, 02:23 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
2. $f(u)=(2^u+2^{-u})^{10}$

Use chain rule,
$10(2^u+2^{-u})'(2^u+2^{-u})^9$
Thus,
$10(\ln 2\cdot 2^u-\ln 2\cdot 2^{-u})(2^u+2^{-u})^9$
Thus,
$10\ln 2(2^u-2^{-u})(2^u+2^{-u})^9$
• July 4th 2006, 02:33 PM
c_323_h
Thank you very much for the quick response...and for solving all the problems.
• July 4th 2006, 02:59 PM
Soroban
Hello, c_323_h!

Quote:

Find the derivative:

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

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

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

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

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

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

Quote:

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

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

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

Quote:

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

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

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

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

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

Quote:

Use the substitution rule:

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

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

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

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

You can finish it now . . .

Quote:

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

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

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

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

• July 4th 2006, 03:02 PM
c_323_h
thanks you. as you can tell i am struggling with this class. is anyone willing to help me with these?

Differentiate:

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

Evaluate the integral:

1. $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}$
• July 4th 2006, 07:04 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h

1. $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}$

Manipulate,
$\int \frac{6}{t-t^2} dt$
As, (since $t\geq 0$) (I factored it)

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

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

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

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

This is a inverse hyperbolic function which you should recognize thus,
$\cosh^{-1} u+C$
Substitute back,
$\cosh^{-1} \sqrt{t}+C$
Thus,
$\left \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}=\cosh^{-1} \sqrt{t} \right|^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}$
• July 4th 2006, 07:19 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
$x^2y'+y^2y'=2x+2yy'$

Could you tell me where you got the $y'$ from on the right side of the equation?
• July 4th 2006, 09:24 PM
Soroban
Hello again, c_323_h!

Quote:

Differentiate:

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

Quote:

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

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

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Since $\cot\alpha = \frac{t}{1} = \frac{adj}{opp}$, $\alpha$ is in this right triangle:
Code:

              *             /β|           /  |1         /α    |       * - - - *           t

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

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

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

No wonder the derivative is 0 . . .