# differentiate

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• August 23rd 2008, 04:07 PM
scuzi
differentiate
My maths assignment is due tomorrow, and i dont understand it. i know how to do it using the secant method, but we are not allowed to use it.

I NEED TO DIFFERENTIATE THIS PROBLEM!!!

tan(230/x)

if anyone can tell me what the differential of this equation is without using a secant method, i will LOVE you forever!!!!
• August 23rd 2008, 04:32 PM
skeeter
$\frac{d}{dx}[\tan{u}] = \sec^2{u} \cdot \frac{du}{dx}$
• August 23rd 2008, 04:59 PM
Spec
$tanu = \frac{sinu}{cosu}$

Let $f(u) = sin u$ and $g(u)=cosu$

Then $\frac{d}{du}\left( \frac{f(u)}{g(u)}\right) = \frac{f'(u)g(u) - f(u)g'(u)}{g(u)^2} = \frac{cosu\cdot cosu - sinu \cdot (-) sinu}{cos^2u}=$ $\frac{cos^2u + sin^2u}{cos^2u} = \frac{1}{cos^2u}$

Now if $u=\frac{230}{x}$, then $\frac{du}{dx} = -\frac{230}{x^2}$

$f'(u(x)) = cosu(x) \cdot \frac{du}{dx} = -\frac{230}{x^2}cos\frac{230}{x^2}$ and $g'(u(x)) = -sinu(x) \cdot \frac{du}{dx} = \frac{230}{x^2}sin\frac{230}{x^2}$

You should be able to handle it from here.
• August 23rd 2008, 05:13 PM
Chop Suey
Quote:

Originally Posted by Spec
$tanu = \frac{sinu}{cosu}$

Let $f(u) = sin u$ and $g(u)=cosu$

Then $\frac{d}{du}\left( \frac{f(u)}{g(u)}\right) = \frac{f'(u)g(u) - f(u)g'(u)}{g(u)^2} = \frac{cosu\cdot cosu - sinu \cdot (-) sinu}{cos^2u}=$ $\frac{cos^2u + sin^2u}{cos^2u} = \frac{1}{cos^2u}$

Now if $u=\frac{230}{x}$, then $\frac{du}{dx} = -\frac{230}{x^2}$

$f'(u(x)) = cosu(x) \cdot \frac{du}{dx} = -\frac{230}{x^2}cos\frac{230}{x^2}$ and $g'(u(x)) = -sinu(x) \cdot \frac{du}{dx} = \frac{230}{x^2}sin\frac{230}{x^2}$

You should be able to handle it from here.

I could be wrong, but I think what he meant by not using the secant method is that he doesn't want to use the definition of the derivative, not the secant trig function. If he/she could only confirm it...
• August 23rd 2008, 06:45 PM
scuzi
confiming
no, im not allowed to use anything to do with secant. we havent been taught it in class, so i cant answer the question using it. :(
• August 23rd 2008, 06:51 PM
Chris L T521
Quote:

Originally Posted by scuzi
My maths assignment is due tomorrow, and i dont understand it. i know how to do it using the secant method, but we are not allowed to use it.

I NEED TO DIFFERENTIATE THIS PROBLEM!!!

tan(230/x)

if anyone can tell me what the differential of this equation is without using a secant method, i will LOVE you forever!!!!

$\frac{d}{\,dx}\tan u=\sec^2u\cdot\frac{\,du}{\,dx}=\frac{1}{\cos^2u}\ cdot\frac{\,du}{\,dx}$

You get away without using a secant term...its in terms of cosine...:D

But is this what you're looking for?

--Chris
• August 23rd 2008, 06:59 PM
scuzi
maybe...
it looks like it. im trying to find the max and min of tan230/x, which i need to diffentiate for. do u think this will work?
• August 23rd 2008, 07:26 PM
Jhevon
Quote:

Originally Posted by scuzi
it looks like it. im trying to find the max and min of tan230/x, which i need to diffentiate for. do u think this will work?

yes, you must find the derivative and set it equal to zero to find the critical points. then you can test these points to find maximums or mins. in this case, there are none since the derivative is never zero
• August 23rd 2008, 07:29 PM
scuzi
so my equation doesnt have a maximum point?
• August 23rd 2008, 07:31 PM
Jhevon
Quote:

Originally Posted by scuzi
so my equation doesnt have a maximum point?

correct. tan(230/x) has no local maxs or mins
• August 23rd 2008, 07:39 PM
scuzi
hmmm okay, that means i got the first equation wrong and its not tan230/x. ummm let me just tell you what im trying to answer.

"Local art gallery wants to know what distance patrons are to stand away from artworks for optimal viewing agle. consider at least 2 scenarios with painting size and hight placements."

so ive made the hieght of the painting 2m (200cm) and the bottome of the artwork 30cm above eye level.

that would make the opposite wall 230cm. and as the adjacent is unknown, i assumed that by using a tan function i could find the optimal angle, and then find the distance. is this the wrong way to go about it?
• August 23rd 2008, 08:03 PM
Spec
You haven't defined what an optimal viewing angle is. Without knowing that, there's no way to tell what an optimal viewing distance is.
• August 23rd 2008, 08:07 PM
scuzi
sorry, im not very good at defining. optimal angle is the largest possible angle.
• August 23rd 2008, 08:17 PM
Spec
And what is that for the human eye? And are you to assume 20/20 vision?
• August 23rd 2008, 08:23 PM
scuzi
well the question doesnt say anything about the vision, so i would assume so
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