# Derivative using the definition

• Flappy
In summary: If you need to integrate and such, just ask me. I don't mind helping.In summary, we used the definition of a derivative to show that f(x)=1+\cos\left(\frac{\pi x}{2}\right) is differentiable at x=2. We evaluated the limit as h approaches 0 and simplified using trigonometric identities until we got the answer of 0. Using rules of differentiation, we can also verify that the derivative at x=2 is 0.
Flappy

## Homework Statement

Use the definition of the derivative to show f is differentiable at x = 2.

f(x) = 1 + cos(pix/2)

## The Attempt at a Solution

I started getting the solution but got stuck at one part.

f`(x) = lim h->0 | [ f(x+h) - f(x) ] / (h)

= lim h->0 | [f(2+h) - f(2) ] / (h)

= lim h -> 0 | [1 + cos(pi(2+h)/2) - 0 ] / (h)

= lim h -> 0 | [1 + cos(2pi+hpi)/2 ] / (h)

I know I am supposed to get the h's to cancel out but I am having trouble getting there

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$$f(x)=1+\cos\left(\frac{\pi x}{2}\right)$$

$$\lim_{h\rightarrow0}=\frac{f(2+h)-f(2)}{h}$$

$$\lim_{h\rightarrow0}=\frac{1+\cos\left[\frac{\pi}{2}(2+h)\right]-(1+\cos \pi)}{h}$$

$$\lim_{h\rightarrow0}=\frac{1+\cos\left({\pi+\frac{\pi h}{2}}\right)}{h}$$

$$\cos{(x+y)}=\cos x\cos y-\sin x\sin y$$

$$\lim_{h\rightarrow0}=\frac{1-cos\left(\frac{\pi h}{2}\right)}{h}\cdot\frac{1+cos\left(\frac{\pi h}{2}\right)}{1+cos\left(\frac{\pi h}{2}\right)}$$

*Difference of Squares

$$\lim_{h\rightarrow0}=\frac{\sin{\left(\frac{\pi h}{2}\right)}}{h}\cdot\frac{\sin{\left(\frac{\pi h}{2}\right)}}{1+cos\left(\frac{\pi h}{2}\right)}\cdot\frac{\frac{\pi}{2}}{\frac{\pi}{2}}$$

Still typing, comp freezes! Refresh every minute if you want.

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Is it, pi/2 or just pi? Reading your last 2 lines, there is a discrepancy.

Oh, I might have mislead with the argument inside cos. It's

$$f(x)=1+\cos\left(\frac{\pi(x)}{2}\right)$$

So what I have so far is good?

Yeah, I see what you did. I was a little confused at first. I forgot to add the 2 under the division on my last 2 lines. Thank you for taking the time by the way.

Hey hold on, my friend needs help with graphing (she's a girl, so :p) But hmm ... basically my LaTeX image isn't showing up. What you want to do is, use a trig identity ... 1-cos^2 x = sin^2 x. Use the fact that sinx\x = 1. So your angle is pi*h/2, so what will you need to do? Think back! I think it will work, I haven't finished working the problem. Been having to type the LaTeX, lol.

Dam, this problem is just nasty.

Practicing latex here. this is after you multiply by the conjugate right?

$$\lim_{h\rightarrow0}= \frac{1 - cos(\frac{{\pi}h}{2})^2}{h(1+cos(\frac{{\pi}h}{2})}$$

Lol, yes. Damn I can't get it to work. There is an easier way. You just got to be neat about it yanno. If you want to make a complex fraction. Put ...

\frac{a}{b}

Then to the side

a=\frac{c}{d}
b=\frac{e}{f}

Then just replace by copy paste :p But obviously I haven't learned my lesson ... argh. And you don't need the brackets if it's just one letter.

\frac a b works great!

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I don't know if I'm on the right track with the problem b/c when I differentiate it doesn't match the definition. I will get 0 by using the definition ... hmm. But of course, we evaluated at 2!

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WOOHOO! It's right, lol. Forgot we evaluated at 2 :p

I was like ... wtf I typed all that up for nothing! Damn I suck.

I believe you are supposed to get 0. Unless I am not following you. I did the derivative using rules and ended up getting 0.

Doing the derivative I got $$-sin\frac{{\pi}x}{2} * \frac{{\pi}}{2}$$

Then by plugging in 2 you would get 0 * pi/2 which is 0.

I'm done typing it up, whew! Btw, there should be a negative somewhere ... not sure where it went, since the derivative of cosx is -sinx. It would be counted wrong even though we get the right answer yanno. Just letting you know. Maybe it disappeared since we evaluated at 2 from the get go :p

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Just a question, where did the pi/2 multiplication come from?

Flappy said:
Just a question, where did the pi/2 multiplication come from?
What is the limit of ... $$\lim_{h\rightarrow0}\frac{\sin{\left(\frac{\pi h}{2}\right)}}{h}$$ ? Definitely not 1 since it's not $$\lim_{h\rightarrow0}\frac{\sin x}{x}=1$$

So I have to multiply both numerator and denominator by pi/2, which is implied multiplication of 1. Basically, the angle and the denominator must be the same!

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I see, i see. Well, thank you for all the help, appreciated.

Flappy said:
I see, i see. Well, thank you for all the help, appreciated.
Anytime :) If you want help like this again, make sure to show work :-] Only reason why I tried my best since I know you tried! Have fun, Calculus gets better.

## What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It tells us how much the output of a function changes when the input changes by a small amount.

## What is the definition of derivative?

The definition of derivative is the limit of the difference quotient as the change in the input approaches zero. In other words, it is the slope of the tangent line to a curve at a specific point.

## How do you find the derivative using the definition?

To find the derivative using the definition, we first need to write out the difference quotient using the given function. Then, we take the limit as the change in the input approaches zero. Finally, we simplify the resulting expression to find the derivative.

## Why is the definition of derivative important?

The definition of derivative is important because it is the basis for the entire concept of calculus. It allows us to find the rate of change of a function at any point, which is crucial in many real-life applications such as physics, economics, and engineering.

## What is the difference between derivative and differentiation?

Derivative and differentiation are often used interchangeably, but there is a subtle difference between the two. Derivative refers to the general concept of finding the rate of change of a function, while differentiation specifically refers to the process of finding the derivative using a set of rules and formulas.

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