Solving e^x Differential and Integral Equations

[madmike159]

What if the proof for differentiating and integrating e^x.

For d/dx (e^x) i used the chin rule and did
u = x
y = e^u
dy/du = ue^u-1
du/dx = 1
dy/du*du/dx = ue^u-1
so you get xe^x-1 but that's not right. I don't even know where to start with intergration.

Can anyone show me? Thanks

(also where are all the symbols have gone? I Don't know how to put them in anymore.)

*edit* Sorry I didn't realize this wasn't home work help section. Can a moderator move it there please.

 

[gabbagabbahey]

Why do you have dy/du=ue^(u-1)?

 

[sutupidmath]

well, there are different ways as how the derivative of e^x is defined. We usually start by the def. of the derivative.

let $$f(x)=a^x$$, be any exponential function, so

$$(a^x)'=\lim_{h\rightarrow 0}\frac{a^{x+h}-a^x}{h}=\lim_{h\rightarrow 0}\frac{a^x(a^h-1)}{h}=a^x\lim_{h\rightarrow 0}\frac{a^h-1}{h}$$

Now, notice that $$\lim_{h\rightarrow 0}\frac{a^h-1}{h}=f'(0)$$ , right? In other words, this is the derivative of the exponential function at the point x=0.

SO, it looks like the derivative of any exponential function at any point, say c, $$f(x)=a^x$$ is simply a constant multiple of its value at that particular point. IN other words, the derivative of any exponential function, and the value of the function at a certian point x are proportional.

Now, it is convinient for us to find an exponential function such that the derivative of that function at x=0 is 1. f'(0)=1.

$$\lim_{h\rightarrow 0}\frac{a^h-1}{h}$$ let $$a^h-1=t=>a^h=t+1$$ also

$$h=log_a=log_a(t+1)$$ notice that when h-->0, t-->0, so;

$$\lim_{h\rightarrow 0}\frac{a^h-1}{h}=\lim_{t\rightarrow 0}\frac{t}{log_a(t+1)}=\lim_{t\rightarrow 0}\frac{1}{log_a(t+1)^{\frac{1}{t}}}=\frac{1}{log_ae}=lna$$


Now as we can see, in order for f'(0)=1, a=e, so $$(e^x)'=e^x$$

 

[madmike159]

Why do you have dy/du=ue^(u-1)?

because u is a value. the power is decreased by one.

Thanks for the help.

 

[gabbagabbahey]

because u is a value. the power is decreased by one.

$$\frac{d}{du} u^e =eu^{e-1}$$

Not

$$\frac{d}{du} e^u=ue^{u-1}$$

To find the derivative of e^x, you will need to use the definition of derivative in terms of limits.

 

[sutupidmath]

because u is a value. the power is decreased by one.

The power rule of differentiating does not apply here, since your power here is not a number, but rather a function/variable.
Like i said in my above post. In my previous post, i explained how why the derivative of exp. functions is what it is.

other ways of doing it, which come indirectly from the def. of derivatives, is

let $$f(x)=a^x$$ then let $$a^x=y=>lny=xlna$$ now let's differentiate implicitly:

$$\frac{y'}{y}=lna=>y'=ylna=>y'=(a^x)'=a^xlna$$ in your case you have $$f(x)=e^x$$ so all you need to do is replace a by e. and youll get what u need.!

 

[HallsofIvy]

I have absolutely no idea what you mean by "u is a value"! Every symbol or expression in mathematics is a value! The question is whether or not u is a variable or a constant. Since you let u= x and x is a variable, so is x.

The derivative of x^a, where a is a constant, with respect to x, is a x^{a-1 but the derivative of ax is ln(a) ax and in the special case that a= 2, ln(e)= 1, that becomes d(ex)/dx= ex.}

 

[madmike159]

Well my working was wrong. I just made u = x (like in the chain rule, e^2x-1 u = 2x-1) and I took 1 off it because when you differentiate the power decreases by 1.