Author Topic: Differentiation (Read 6610 times) Tweet Share

d0minicz · « **on:** April 27, 2009, 04:08:32 pm »

Let $f: (-\frac {\pi}{2} , \frac {\pi}{2}) \rightarrow R , f(x) = tanx- 8sinx$ .
a)Find :
i) the stationary points on the graph $y= f(x)$
ii) the nature of each of the stationary points.

thanks =]

TrueTears · « **Reply #1 on:** April 27, 2009, 04:25:05 pm »

$f'(x) = sec^2(x) - 8cos(x)$

now this equals $\frac{1-8cos^3(x)}{cos^2(x)}$ after letting $sec^2(x) = \frac{1}{cos^2(x)}$ and common denominating etc.

now we require $f'(x) = 0$

$\therefore$ $\frac{1-8cos^3(x)}{cos^2(x)}$ $= 0$

solving for x yields $\frac{-\pi}{3}$ or $\frac{\pi}{3}$ within the domain given (I assume you can solve this since you have already done circular functions)

subbing in x = $\frac{-\pi}{3}$ or $\frac{\pi}{3}$ in f(x) yields $f(\frac{\pi}{3}) = -3\sqrt{3}$ and $f(\frac{-\pi}{3}) = 3\sqrt{3}$

stationary points are $(\frac{\pi}{3} , -3\sqrt{3})$ and $(\frac{-\pi}{3} , 3\sqrt{3})$

TrueTears · « **Reply #2 on:** April 27, 2009, 04:31:51 pm »

now for part 2 we need to find f''(x) (second derivative)

now deriving the $\frac{1-8cos^3(x)}{cos^2(x)}$ yields $\frac{2sin(x)(4cos^3(x)+1)}{cos^3(x)}$ (Quotient rule)

subbing in $x = \frac{-\pi}{3}$ in $\frac{2sin(x)(4cos^3(x)+1)}{cos^3(x)}$ yields $-12\sqrt{3}$ which is < 0

now we know if f''(x) < 0 and f'(x) = 0 then it is a local maximum (check your essentials book there is a huge table for this)

now subbing in $x = \frac{\pi}{3}$ in $\frac{2sin(x)(4cos^3(x)+1)}{cos^3(x)}$ yields $12\sqrt{3}$ which is > 0

now we know if f''(x) > 0 and f'(x) = 0 then it is a local minimum (again check the book)

NE2000 · « **Reply #3 on:** April 27, 2009, 04:41:33 pm »

Quote from: TrueTears on April 27, 2009, 04:31:51 pm

now for part 2 we need to find f''(x) (second derivative)

now deriving the $\frac{1-8cos^3(x)}{cos^2(x)}$ yields $\frac{2sin(x)(4cos^3(x)+1)}{cos^3(x)}$ (Quotient rule)

subbing in $x = \frac{-\pi}{3}$ in $\frac{2sin(x)(4cos^3(x)+1)}{cos^3(x)}$ yields $12\sqrt{3}$ which is > 0

now we know if f''(x) > 0 and f'(x) = 0 then it is a local maximum (check your essentials book there is a huge table for this)

now subbing in $x = \frac{\pi}{3}$ in $\frac{2sin(x)(4cos^3(x)+1)}{cos^3(x)}$ yields $-12\sqrt{3}$ which is < 0

now we know if f''(x) < 0 and f'(x) = 0 then it is a local minimum (again check the book)

If f''(x) > 0 then the gradient is increasing and therefore it is a local minimum isn't it?

Eg. f(x) = x^2 (a minimum)
f'(x) = 2x
f''(x) = 2

f''(x) > 0 then there is a local minimum

If f''(x) < 0 then it is a local maximum

d0minicz · « **Reply #4 on:** April 27, 2009, 05:12:42 pm »

Let $f: (-\frac {\pi}{4} , \frac {\pi}{4}) \rightarrow R , f(x) = tan(2x)$ . A tangent to the graph of $y=f(x)$ where $x=a$ makes an angle of $70^o$ with the positive direction of the x axis. Find the possible values of $a$ .

thankz

Flaming_Arrow · « **Reply #5 on:** April 27, 2009, 05:28:34 pm »

differntiate tan(2x) and make it equal to tan(70)

TrueTears · « **Reply #6 on:** April 27, 2009, 05:41:58 pm »

dom you might wanna take a look here http://vcenotes.com/forum/index.php/topic,11962.0.html

and also FA i think its equal $tan(70^o)$ which is $tan(\frac{7\pi}{18})$ (Must convert to radian), think dekoyl had the same problem.

Over9000 · « **Reply #7 on:** April 27, 2009, 05:57:39 pm »

so $tan(\frac{7\pi}{18}) = 2sec^{2}(2x)$
$\frac{tan(\frac{7\pi}{18})}{2} = sec^{2}(2x)$
$\frac{tan(\frac{7\pi}{18})}{2} = \frac{1}{cos^{2}(2x)}$
flip fractions
$\frac{2}{tan(\frac{7\pi}{18})} = cos^{2}(2x)$
square both sides
$\frac{\sqrt{2}}{\sqrt{tan(\frac{7\pi}{18})}} = cos(2x)$
rationalise
$\pm\frac{\sqrt{2tan\frac{7\pi}{18}}}{tan\frac{7\pi}{18}} = cos2x$
u should be able to do from here

TrueTears · « **Reply #8 on:** April 27, 2009, 06:00:16 pm »

nice solution over9000

d0minicz · « **Reply #9 on:** April 27, 2009, 06:08:19 pm »

Let $f(x) = sec (\frac {x}{4})$ .
a) Find f'(x) , which is $\frac {1}{4} sin (\frac {x}{4})sec^2 (\frac {x}{4})$
b) Find f'(pi), which is $\frac {\sqrt 2}{4}$
c) Find the equation of the tangent of y= f(x) at the point where $x= \pi$
thx

Flaming_Arrow · « **Reply #10 on:** April 27, 2009, 06:17:24 pm »

GerrySly · « **Reply #11 on:** April 27, 2009, 06:24:41 pm »

Well you have the gradient at that tangent, then just use the standard form $y=mx+c$

$\begin{align*} y=\frac{\sqrt{2}}{4}x+c \end{align*}$

When $x=\pi$

$\begin{align*}y&=\sec\left ( \frac{\pi}{4} \right )\\ &=\sqrt{2} \end{align*}$

Therefore

$\begin{align*} \sqrt{2}&=\frac{\sqrt{2}}{4}\pi+c\\ c&=\sqrt{2}-\frac{\sqrt{2}}{4}\pi\\ &=\frac{4\sqrt{2}-\sqrt{2}\pi}{4}\\ &=\frac{\sqrt{2}(4-\pi)}{4} \end{align*}$

So we have

$\begin{align*} y&=\frac{\sqrt{2}}{4}x+\frac{\sqrt{2}(4-\pi)}{4}\\ &=\frac{\sqrt{2}}{4}(x+4-\pi) \end{align*}$

Edit: Damn it, I got beaten to it again

d0minicz · « **Reply #12 on:** May 09, 2009, 12:23:08 pm »

Find f''(0) if f(x) is equal to:
a) $\sqrt {1-x^2}$
b) $tan^{-1} (\frac {1}{x-1})$
need steps , thanks =)

TrueTears · « **Reply #13 on:** May 09, 2009, 01:22:02 pm »

$\frac{d}{dx}\sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} = \frac{1}{2}(1-x^2)^{\frac{-1}{2}} \times -2x = \frac{-x}{\sqrt{1-x^2}}$

$\frac{d}{dx}\frac{-x}{\sqrt{1-x^2}} = \frac{-1(\sqrt{1-x^2}) - \frac{-x}{\sqrt{1-x^2}}(-x)}{1-x^2}$

The numerator becomes: $-\sqrt{1-x^2} - \frac{x}{\sqrt{1-x^2}} = \frac{-(1-x^2) - x}{\sqrt{1-x^2}} = \frac{x^2-x-1}{\sqrt{1-x^2}}$

Subbing this back into the fraction: $\frac{x^2-x-1}{\sqrt{1-x^2}(1-x^2)}$

Now subbing in 0 for x yields $f"(0) = -1$

TrueTears · « **Reply #14 on:** May 09, 2009, 01:34:42 pm »

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