Can anyone teach me how to differentiate this one?y=x(13x)^5

Can anyone teach me how to differentiate this one?y=x(13x)^5
Ooh, it's been a while since I've done any differentiation, but I'll give it a go.
This differentiation will require a combination of the product rule and the chain rule, so:
y = x(13x)<sup>5</sup>
Suppose u = x and v = (13x)<sup>5</sup>
y' = uv' + u'v
u' = 1, that's easy. v' is a bit more complicated, and requires the chain rule.
v = (13x)<sup>5</sup>
Suppose that v = w<sup>5</sup> and w = (13x)
dv/dx = dw/dx x dv/dw
dw/dx = 3 and dv/dw = 5u<sup>4</sup>
Hence:
v' = 3(5(13x)<sup>4</sup>) = 15(13x)<sup>4</sup>
Put that into the original expression:
y' = uv' + u'v = x(15(13x)<sup>4</sup> + (13x)<sup>5</sup>
y' = (13x)<sup>5</sup>  15x(13x)<sup>4</sup>
Probably simplifies to something a lot nicer, but I can't really be bothered at the moment, sorry :P I *think* I've done it right, but it's been a while so I can't really be sure anymore
Got the same thing Matt
It simplifies to (1  3 x)<sup>4</sup> (1  18 x)
Yeah I got that simplification too. Couldn't be bothered to edit the post though, I didn't consider it an amazing improvement, haha.
I tutor engineers this stuff so the simplification comes naturally.
Cool. This forum gives me the odd good excuse to keep up on my maths Since I'm a materials chemist these days I don't have much use for pure maths anymore :P
So you're a lab jockey  not for me!
Welcome to the forum
Thanks I'm starting to like this place
how does dv/dw become 5u<sup>4</sup>Originally Posted by Matt Lacey
what is 'u' anyhow?
Its a typo wallaby
Should read
dv/dx = dw/dx x dv/dw
dw/dx = 3 and dv/dw = 5w<sup>4</sup>
where w = (1  3x)
Use the formulae d(u*v)=u*d(v)+v*d(u) and d(x^n)=n*x^(n1), n is constant in your question and simplify to get final result. Afterall, its your homework and I need not do intermediate steps for you
Sorry, my badOriginally Posted by river_rat
I had it stuck in my head to use dy/dx = du/dx x dy/du, temporarily forgot I was using different symbols. Oooops...
thanks for clearing that up, i myself am new to differentiation so it is easy for me to get confused.Originally Posted by Matt Lacey
can anyone tell me what are the uses of differentiation?>
You bastard Matt  beat me to it :)
The differential of a function gives you a formula for the gradient of that function at any x (or whatever the independent variables are).Originally Posted by cheenaik
so, if y = x^2, dy/dx = 2x which means at x=3 the gradient of x^2 is 6.
In general, differentation can be described as the 'rate of change of something with respect to something else'.
I'm sure others can give a slighter more rigorous reply
Differentiation is important as isOriginally Posted by cheenaik
 Gives possible local maximum and minimimum values for a "nice" function (optimisation and mathematical programming problems)
 Is the definition for velocity and acceleration so forms the basis for dynamics and kinematics
 Derivatives are used to give approximations to rather nasty looking functions (ie. Taylor expansions)
 Derivitives are used to solve equations numerically (ie Newtons method)
 Most importantly, as derivitives give "rate of change" information they form the basic ingrediants for Differential Equations which are ubiquitous in the sciences and one of the main friving forces for modern mathematical research.
Thats just off the top of my head, there are a few more reasons  anybody got anything else to add?
The equation isOriginally Posted by cheenaik
You can use what's known as the "chain rule" to differentiate this function. Firstly however, you must learn a less sophisticated rule known as the "power rule."
Assume we have a function , in order to find the derivative of it, we must plug it into the the difference quotient, which is the following formula:
. From this, we derive the following formula (note that we can't pass to limit yet, as our output is an undefined quantity, therefore we must make some cancellations first):
Now, we can use the binomial theorem to simplify the numerator to the form , where are the remaining terms within the expansion of the binomial. Now, we can make our cancellations: and to begin with cancel one another out.
From that, we have the remaining expression:
Based on the properties of reciporicals, the above expression simplifies to
Now we can simply pass to the limit, or , which is based on the fact that all of the remaining terms have a coefficient of and the change in x tends to zero. What this means is that .
Here is where the chain rule comes into play: it states that if we have a compound function , than its derivative is
Than the derivative of the function is equal to
« cubic equation  Geometry question » 