# Thread: Differential equation

1. ## Differential equation

Hi All,
Here is a diffential equation i need your help in:

Differentiate y w.r.t. x where

Y = e (log tan x^4)^2

Thanks
Arun

2. I don't know if you can call this a D.E. technically, but let ThePerfectHacker, CaptainBlack, or some other knowledgable person clarify that for you. I have the idea of what you want done.

Now I'm assuming that when you wrote "e" you meant e to the power of the rest of your problem. If it's e times the rest, than treat e as a constant. I'm accustomed to seeing log as synonymous with ln, but I'll assume you meant log base 10.

Y = e^[(log[tan(x^4)])^2]

dY/dx = ???

Well let's start with the outside function, e^[f(x)]. It's derivative is e^[f(x)]*f'(x). Same thing here, it's just f'(x) is a bit messy.

So start with dY/dx = e^[f(x)]*f'(x), where f(x) = [log(tan[x^4])]^2.

Now for f'(x). You have two inner terms all squared so we use the power rule. Call log(tan[x^4]) g(x). So we have g(x)^2. And the derivative of that is 2g(x)*g'(x). So let's find g'(x).

The outer function here is log. In case you didn't know, the rule for d(log f[x])/dy is it's equal to [ln(x)/ln(10)] * f'(x). So let'd do it!

Call log[tan(x^4)] z(x). Then z'(x) = [(ln(tan[x^4]) / ln(10)] * z'(x). NOW for z'(x). Does this ever end?!

I'll just do this in one step. I'm sure you can follow. z'(x) = sec^2(x^4) * 4x^3.

Done! Now all we have to do is put that mess together. I get...

dY/dx = (e^[(log[tan(x^4)])^2] * 8log[tan(x^4)]*ln[tan(x^4)]*sec^2(x^4)*x^3) / ln(10)

I think that's the final answer. If I made a slight error, the concept is still right. No matter how complicated, always start from the outside and work your way in!

Jameson