Hyperbolic Functions - Math Course, Self-Study

Hi Sulexk,
I have done some reading on the chain rule, there are many variations for the type of integer or type, using this...

http://www.sosmath.com/calculus/diff/der04/der04.html

I think it is a matter of practice, I will try to solve some of your questions as soon as possible, using the support given and the website, it is all a matter of practice =).

Thanks.

Hi,

there is a unity that exists between all of those equations listed in that website. The commonality is, that the derivative of all can be found using the chain rule, it simply depends how you use it. However for each of them, one can find out dy/dx by using the substitution dy/dx=dy/du * du/dx.
Yes, practice indeed is very important
That substitution is awesome! (Substitutions like those above are tremendously useful in C4, especially when you come across rate of change problems!!) So having a brief idea of them now will greatly help.

Hi Sulexk,

So; ques 1: y= ln(4x+5)

Using the derivative : (dy/du) * (du/dx) = dy/dx

(1/4x+5) * (4x+5/4x) = 1/4

Assuming y = 1 based on 1/u == y = ln(u)

== means equal to

Is this correct or does this need to be done differently?

Thanks =).

Hi, good attempt my friend

So, we know that dy/dx= (dy/du)*(du/dx), and that y=ln(4x+5)
so we have a function of a function again, i.e. 4x+5 is one function, then taking
"ln" of this is another function. So let us rewrite this as y=ln(u) where u=4x+5
then du/dx = 4 and as y=ln(u) then dy/du = 1/u, so:

dy/dx = (dy/du)*(du/dx) = (1/u) * 4 = 4/u
since u=4x+5
then dy/dx = 4/u=4/(4x+5)
So its the whole function of a function reasoning!

Hope that helps!

Hi Sulexk,

Thanks , at first I was like "whaaa" , thanks to your help I am like...



haha, thanks.