Solving calculus midterm without tan addition formula

[toosm:)ey]

Hello everyone!

I have a calculus midterm next week and I need some help solving old midterm questions. I'm a first year calculus student so they're not hard, but I can't figure them out!

I am told to solve without using the tan addition formula.

 

[toosm:)ey]

problem number 2

 

[arildno]

Since:
$$tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}$$
the question merely asks you to calculate the derivative of tan(x) at $$x=\frac{\pi}{6}$$

 

[arildno]

For your other questions, show some of your own work on these.

 

[toosm:)ey]

Sure, but how do I use the code thing? Or is there an easy way to type out the work I have?

 

[arildno]

Sure, but how do I use the code thing? Or is there an easy way to type out the work I have?

Click on a given "LATEX" code to see how you generate it. There's a sticky in General Physics which tells you a bit about it.

 

[toosm:)ey]

$$\lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(h+(\pi/6))}{cos(h+(\pi/6))}} - \frac{1}{\sqrt3}})$$

$$\lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(x)cos(\pi/6) + cos(x)sin(\pi/6)}{cos(x)cos(\pi/6) + sin(x)sin(\pi/6)}} - \frac{1}{\sqrt3}})$$

$$\lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(x)(\sqrt3/2) + cos(x)(1/2)}{cos(x)(\sqrt3/2) + sin(x)(1/2)}} - \frac{1}{\sqrt3}})$$

$$\lim_{h\rightarrow0} \frac{1}{h}} ({\frac{\sqrt3sin(x) + cos(x)}{{\sqrt3cos(x) +sin(x)}} - \frac{1}{\sqrt3}})$$

$$\lim_{h\rightarrow0} \frac{1}{h}} (\frac{2sin(x)}{3cos(x) + \sqrt3sin(x)})$$

Wow, that latex is hard to code... I think I have it now though.

This is asfar as I can go with this problem, from here, I am confused.

 

[allistair]

you're making it way to diffcult, lim(h->0) 1/h * (tan(Pi/6 + h)-1/sqrt(3)) is also lim(h->0) 1/h * (tan(Pi/6 + h)-tan(Pi/6)), lim(h->0) 1/h * f(a+h)-f(a) is the differential of f in a, so you really just need to calculate tan's differential in Pi/6

 

[toosm:)ey]

Okay, that makes sense and therefore:

$$\lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \frac{1}{\sqrt3}})$$

equals

$$\lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \tan(\frac{\pi}{6}}))$$

$$\frac{d}{dx}}\tan(x) = \sec^2(x)$$

$$\sec^2(\frac{\pi}{6}}) = \frac{4}{3}}$$

and


$$\lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \frac{1}{\sqrt3}}) = \frac{4}{3}}$$

Is this true?

 

[allistair]

yeah, that's what i meant