defferentiaton

[lakshmi]

can anybody differentiate this to simplest form
y=4/3+5cosx

[arildno]

What about trying it out yourself?
What is your specific problem with this?

[HallsofIvy]

Do you know the derivative of a constant?

Do you know the derivative of cos(x)?

[nix]

for 5cos(x) product rule may be useful

[HallsofIvy]

By the way, is that (4/3)+ 5cos(x) or 4/(3+5cos(x))?

[Kurdt]

nix: The product rule does not need to be applied as the 5 looks to be independant of x.

[Electro]

Hello,
I have a limit to solve which in fact involves differentation (L'Hospital's rule)

Lim as x->inf. (xe^[1/x]-x)
I solved it but I am not quite sure. It can't be solved immediately by substitution because it becomes an undet. form. So applying the Hospital Rule I found Lim as x->inf (e^[1/x] - [(e^{1/x})/x] - 1)
Then it gives e^0 -0-1 = 1-1=0
I don't know now if it's correct because I don't even have a TI calc. to check the answer.
Thanx

[faust9]

Your answering a question that is a month old and that isn't the correct solution either.

y=\frac{4}{3+5\cos{x}}


You can use the quotient rule or change the denomenator to a negative power

Quotient rule:

given a function f(x)

f(x)=\frac{g(x)}{h(x)}


Then the quotient rule would be:

f^\prime (x)=\frac{h(x)g^\prime (x)-h^\prime (x)g(x)}{(h(x))^2}


if we let the numerator be g(x) and the denomenator be h(x) we get:

g(x)=4



g^\prime (x)=0



h(x)=3+5\cos{x}



h^\prime (x)=-5\sin{x}


Plugging the individual functions into the quotient rule yields:

f^\prime (x)=\frac{(3+5\cos{x})(0)-(-5\sin{x})(4)}{(3+5\cos{x})^2}


simplify:

f^\prime (x)=\frac{20\sin{x}}{(3+5\cos{x})^2}


If you look at the problem as:

y=\frac{4}{3}+5\cos{x}


you get:

y^\prime=-5\sin{x}



[edit]transposed 3 and 4...fixed

[HallsofIvy]

Your answering a question that is a month old and that isn't the correct solution either.
y= \frac{4}{3+5cos x}
How did you decide that that was the problem? I would think that y= 4/3+ 5 cos x would be more reasonably interpreted as y= (4/3)+ 5 cos x. I asked lakshmi earlier which was intended and got no answer.

[faust9]

Your answering a question that is a month old and that isn't the correct solution either.
y= \frac{4}{3+5cos x}
How did you decide that that was the problem? I would think that y= 4/3+ 5 cos x would be more reasonably interpreted as y= (4/3)+ 5 cos x. I asked lakshmi earlier which was intended and got no answer.

I answered both possible forms of the question.

[edit]Also, KnowledgeIsPower you have a misconception about the generel form. the derivative x^n only applies to x^n. You can't apply this to the transcendental functions because each has its own rule. For example: the derivative \ln x is not 1/\ln x as x^n would suggest.

[edit][edit] The response I was addressing seems to have disappeared anyway.

[KnowledgeIsPower]

Ok thanks for the information. I just studied a number of new trig content and thought i'd jump right in - though i hadn't done differentiation/integration on trigonomic functions.
My mistake, sorry.

But what prevents you from just classing sinx as 'y' and differentiating that way?

[faust9]

You can say y=sin x but you have to use the chain rule to get dy/dx.

Here, given: y=sin x and you want to find dy/dx.

let A=sin x

y=A
dy=dA
dy/dA=1

Well that's not what we wnat we are trying to find dy/dx


\frac{dy}{dx}=\frac{dy}{dA}\frac{dA}{dx}


The above works just like algebra. The dA's will cancel to give dy/dx; however, you must find dA/dx to use the chain rule.

A=sin x
dA=cos xdx
dA/dx=cos x

thus dy/dx=(1)(cos x)
or dy/dx=cos x

back to the original problem: y=sin x
You'll learn this soon enough, but the derivative of sin x is cos x so and the derivative of cos x is -sin x.

thus if we find the derivative of y=sin x we get dy/dx=cos x

I'd liken the process to fractional addition. You can't simply add 1/3 and 1/2 together. You have to first find a common denominator. Here too we can't simply say sin x=A and get the same result as y=sin x. We still need to do the intermediate steps to convert dA/dx and dy/dA to dy/dx.

You'll soon learn that there are 20 to 30 common derivatives you'll need to memorize.

Hope this helped.

[KnowledgeIsPower]

Thanks a lot for the explanation. That was very interesting and informative. Fools rush in..., guess i shouldn't jump ahead.
It's amazing that you people will help those less knowledgable out for free. It's a pity there's no way to return the favour.
Thanks again.