Derivatives: first deriv. of f(x) = sq. root of 4 sin x + 2

[Declassified]

I just started calculus. I was no good at pre calc and i'm freaking out. Just got a bunch of hw due tomorrow.
I need to find the first derivative of f(x)=sq. root of 4sinx+2
first derivatives are usually pretty easy for me but with a square root involved i'm not sure how to takle this one...
please help me!

 

[mmm4444bot]

Re: Derivatives

I was no good at pre calc ...

Hi Declassified:

Your statement above serves as a red flag, for us AND for you.

You did not post any work, so you've forced us to GUESS why you're stuck. If we don't know what KIND of help you are attempting to get, then it's like the blind leading the blind.

Why are you stuck?

Does it help you to differentiate the expression that defines function f if you see it in the following equivalent form?

[4 sin(x) + 2][sup:1ksavj7y]½[/sup:1ksavj7y]

If NOT, then do you know how to differentiate x[sup:1ksavj7y]½[/sup:1ksavj7y] ?

If NOT, then please explain to us WHY you're stuck.

Cheers,

~ Mark

 

[Declassified]

Re: Derivatives

i get that i can view the expression as (4sinx+2)^1/2 but even so finding the first derivative of that doesnt make sense.
in a normal equation such as x^2 the first derivative is simple=2x. or even first derivative of cosx=sinx. but with the above expression i cant use the simple calculation or even something like the product rule cuz its not multiplying. or the quotient rule cuz there is no division. so how do i even begin to calculate a first derivative with nothing to work off of. i dont know how else to explain WHY or HOW i am confused. that is why i HATE math this year. and this is why i DONT ever ask questions in class. cuz I DONT KNOW

 

[mmm4444bot]

Re: Derivatives

... i cant use the simple calculation ...

Which simple calculation were you thinking about when you wrote this?

There is a simple procedure that involves only two steps that I would use to find the first derivative of x^2.

I'm wondering if that procedure is the same thing that you're thinking about when you say, "simple calculation".

I use this same procedure to differentiate x^(1/2)

I also use it as a basic "template" when I think about differentiating [expression]^(1/2).

So, please tell me which simple calculation you were thinking about when you wrote those words.

Also, please tell me what the CHAIN RULE means to you.

Here's an example:

f(x) = [ 6 ln(x) - 3 ]^(1/2)

To determine the first derivative of this function, I know that I need to do the following.

Multiply [6 ln(x) - 3] by 1/2

Subtract 1 from the exponent

AND, since the expression being raised to the 1/2 power is ALSO A FUNCTION of x (in-and-of itself), the chain rule tells me that I also have to multiply by the first derivative of 6 ln(x) - 3.

(1/2) * [6 ln(x) - 3]^(1/2 - 1) * 6/x

I would finish by using introductory algebra to simplify this result.

\(\displaystyle f'(x) = \frac{\sqrt{3}}{x \sqrt{2\;ln(x) - 1}}\)

The same strategy is used with your exercise.

i don't know how ... to explain [why I'm confused] ... [and] this is why i HATE math this year ... and ... [this is why] i DONT ever ask questions in class

Maybe a person who fails precalculus is not ready to study calculus. I'm NOT saying that you are a failure; I'm just stating this as FOOD FOR THOUGHT.

If you think that there is some sort of QUICK FIX that we can provide to grease the wheels of learning calculus for you, then you think wrongly.

If you do not know how to design a question to obtain information you need OR if you don't want to think about it, then I strongly feel that you should speak with your instructor because these conditions indicate that you are currently too far away from understanding for us to help in the absense of specific questions.

Anyway, please explain what you're thinking about when you write "simple calculation".

~ Mark

 

[skeeter]

Re: Derivatives

i get that i can view the expression as (4sinx+2)^1/2 but even so finding the first derivative of that doesnt make sense.
in a normal equation such as x^2 the first derivative is simple=2x. or even first derivative of cosx=sinx. but with the above expression i cant use the simple calculation or even something like the product rule cuz its not multiplying. or the quotient rule cuz there is no division. so how do i even begin to calculate a first derivative with nothing to work off of. ...

Declassified ... has your class covered the chain rule for derivatives yet?