1. ## 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!!!!

2. $\displaystyle \frac{d}{dx}[\tan{u}] = \sec^2{u} \cdot \frac{du}{dx}$

3. $\displaystyle tanu = \frac{sinu}{cosu}$

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

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

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

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

You should be able to handle it from here.

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

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

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

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

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

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...

5. ## 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.

6. 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!!!!
$\displaystyle \frac{d}{\,dx}\tan u=\sec^2u\cdot\frac{\,du}{\,dx}=\frac{1}{\cos^2u}\ cdot\frac{\,du}{\,dx}$

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

But is this what you're looking for?

--Chris

7. ## 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?

8. 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

9. so my equation doesnt have a maximum point?

10. Originally Posted by scuzi
so my equation doesnt have a maximum point?
correct. tan(230/x) has no local maxs or mins

11. 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?

12. 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.

13. sorry, im not very good at defining. optimal angle is the largest possible angle.

14. And what is that for the human eye? And are you to assume 20/20 vision?

15. well the question doesnt say anything about the vision, so i would assume so

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