Questions about derivatives and etc

[kaitamasaki]

first.. its about the quotient rule

say we are findnig the derivative of something such as x / (x-1)^2
for the derivative of the denominator, do we treat the denominator's exponent as 2 or -2, since its a denominator? (confused with the perspective and approach you take)



today i was exposted to some very confusing stuff

TAKE DERIVATIVE

1. y = ln5x
y' = 1/5x (5x) = 1/x

chain rule was applied

but how come u take out 5x?!. u only take out values that make the derivative an "impure" x or whatever, shouldn't u just take out the 5?!?

whats being added on to 1/x is 5 isn't it? , not 5x

if a 5x was added on, that would be 1 / 5x^2

this doesn't make sense


also, today we took higher ordre derivatives

y=sin2x
y' = 2cos2x i know deriv of sin is cos, but why is there a new coefficent of 2 in front?
y'' = 2(-sin2x)(2) = -4sin2x <--- what the heck? okay so that kind of make sense, but i don't see chain rule applied to y'
y''' = -4cos2x if the transition of cos to sin is multiplied by -2, then why does the coefficient say the same when sin -> cos? arghhhh


note that these are all correct for sure.. just that i don't understand them



even more confusing is the d/dx thing

sometimes its like dy/dx , d/dx, whatever, i don't get it at all,, i know its derivative something respect to x, but i just don't understand how it works

 

[Pengwuino]

Well the probably is that you arent multiplying ln(x) by 5 like you would multiply x^2 by 5.

As for the sin and cos; you are suppose to use u substitution and turn it into f(x)=sin(u) du; u = 2x, du = 2. You continue to do that for every derivative.

 

[Lyuokdea]

When you use the quotient rule, you would use the positive two, for your example:

f(x) = \frac{x}{(x-1)^2}

Set f(x) = x and g(x) = (x-1)^2

f&#039;(x) = \frac{f&#039;(x)g(x) - f(x)g&#039;(x)}{g(x)^2} = \frac{1(x-1)^2 - 2x(x-1)}{(x-1)^4}

To check that this is true, you can always move the expression in the denominator into the numerator and then use the product rule, you can do this to check the answer I've given.

For the natural log problem, you don't take out a 5x you take out the derivative of 5*x which is 5, thus you get 1/5*x * 5 = 1/x.

In your higher order derivitives problem, again the chain rule specifies that in order to find

\frac{d}{dx} f(u) = \frac{df}{du} \frac{du}{dx}

Thus you find for sin(2x) that you can temporarily substitute in u = 2x

\frac{d}{dx} sin(2x) = \frac{d}{du} [sin(u)] \frac{du}{dx}

from above we have u = 2x so du/dx = 2 so we have (substituting 2x back in for u at the end) :

\frac{d}{du} [sin(u)] \frac{du}{dx} u = cos(u) * 2 = 2cos(2x)

Hope this helps!

~Lyuokdea

 

[cepheid]

first.. its about the quotient rule

say we are findnig the derivative of something such as x / (x-1)^2
for the derivative of the denominator, do we treat the denominator's exponent as 2 or -2,

The denominator's exponent IS 2. That's GIVEN. If you would like to move the denominator up into the numerator, then you would change the exponent to -2 (think about it). Then you can use the product rule instead of the quotient rule when differentiating. Does that make sense?

TAKE DERIVATIVE

1. y = ln5x
y' = 1/5x (5x) = 1/x

chain rule was applied

The 5x I highlighted in red is INCORRECT. You are right that it should indeed be a 5:

let u = 5x
then y = ln u

chain rule:

\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{1}{u} \frac{d}{dx}(5x) = \frac{1}{5x}(5) = \frac{1}{x}

if a 5x was added on, that would be 1 / 5x^2
this doesn't make sense

No actually, it would come out to 1, which still doesn't make sense, because it is wrong

also, today we took higher ordre derivatives

y=sin2x
y' = 2cos2x i know deriv of sin is cos, but why is there a new coefficent of 2 in front?

The derivative of sinx is cosx, but here you have a composition of functions, so you must apply the chain rule. Try it by first making the substitution u = 2x, and proceed in a similar way to the ln example I did above. That helps to see why you multiply by 2 when you are first getting used to differentiating. Eventually you'll be able to see that by inspection.

y'' = 2(-sin2x)(2) = -4sin2x <--- what the heck? okay so that kind of make sense, but i don't see chain rule applied to y'
y''' = -4cos2x if the transition of cos to sin is multiplied by -2, then why does the coefficient say the same when sin -> cos? arghhhh

I don't really understand what you are saying. As Pengwuino pointed out, you're just differenting multiple times, following the same rules consistently. Work it out with pencil and paper to verify it instead of just staring at it and deciding it doesn't make sense. That's the only way to do it when you can't follow the steps just by reading them.

 

[cepheid]

Thus you find for sin(2x) that you can temporarily substitute in u = 2x

\frac{d}{dx} sin(2x) = \frac{d}{du} [sin(u)] \frac{du}{dx} u

This is wrong. There shouldn't be any u at the end of that last term.

\frac{d}{du} [sin(u)] \frac{du}{dx} u = cos(u) * 2 = 2cos(2x) 2

The last 2 also should not be there.

 

[Lyuokdea]

thanks, made some typos. They're fixed now to avoid confusion.

~Lyuokdea