Derivative Confusion

HellishHeat

I'm just getting back into this maths lark after a long hiatus and I'm struggling with derivatives.

f(x) = e^x.. f'(x) = e^x. but e is just a constant (2.71...) .so why is the derivative not 0? I find this confusing.. if f(x) = 2.7 .. is f'(x) = 0?

Also,

I think I understand derivatives now to an extent.. I understand power rule, product rule, chain rule. I'm a bit unsure of partial derivatives but I understand the thrust of it.. or at least I thought I did.

f(x,y) = 1/xy fx(x,y) = 1/(x^2)y

I don't get this.. with respect of x means that y can be taken as some arbitrary constant so, for the sake of getting the derivative of 1/xy with resp. of x, it's the same as saying get the derivative of 1/x2 (if y is just come constant, I'll use 2).. which is (x2)^-1 .. I don't see how the derivative of this could be 1/(x^2)2. Can someone enlighten me please?

I'm checking my answers here: http://www.numberempire.com/derivatives.php

MathsManiac

HellishHeat wrote: »

I'm just getting back into this maths lark after a long hiatus and I'm struggling with derivatives.

f(x) = e^x.. f'(x) = e^x. but e is just a constant (2.71...) .so why is the derivative not 0? I find this confusing.. if f(x) = 2.7 .. is f'(x) = 0?

Yes, e is just a constant, but e^x is not a constant. It varies with x.

If f(x)=e, then f'(x)=0. But if f(x)=e^x, f'(x) is not 0.

If you look at the graph of any number to the power of x, (such as 2^x), you'll see that the larger x gets, the steeper the graph gets. Its rate of growth is, in fact, directly proportional to the value of the function. And the derivative measures the rate at which a function is growing as x increases.

So, for example, if f(x)=2^x, then f'(x) is a constant multiple of 2^x.
Similarly, if f(x)=3^x, then f'(x) is a constant multiple of 3^x.

The special number e has the property that this multiple turns out to be 1, so that if f(x)=e^x, then f'(x)=e^x as well.

HellishHeat

Ok.. so f(x) = 4^x ; f'(x) = 4x.. similarly f(x) = e^x; f'(x) = ex = e^x ?

MathsManiac

HellishHeat wrote: »

f(x,y) = 1/xy fx(x,y) = 1/(x^2)y

I don't get this.. with respect of x means that y can be taken as some arbitrary constant so, for the sake of getting the derivative of 1/xy with resp. of x, it's the same as saying get the derivative of 1/x2 (if y is just come constant, I'll use 2).. which is (x2)^-1 .. I don't see how the derivative of this could be 1/(x^2)2. Can someone enlighten me please?

I think you're missing a minus sign, by the way.

Yes, when finding the partial derivative with respect to x, you treat y as though it were a constant. I'm assuming that in the function you gave, both the x and the y are in the denominator of the fraction. That is, I assume you mean 1/(xy).

First, note that the derivative of 1/x is -1/(x^2). You can do that with the power rule: 1/x is the same as x^(-1), so that its derivative is (-1)x^(-2). If you're treating y as a constant, then 1/(xy) is just the same as 1/x multiplied by the constant 1/y. So this constant simply remains as it is. (The derivative of k*f(x) is k*f'(x).) So the answer is -1/((x^2)*y).

Alternatively, you can think of the function as (xy)^-1. Differentiating wrt x, and treating y as a constant gives (-1)(xy)^(-2)*y, using the chain rule. This simplifies to the same answer as before.

MathsManiac

HellishHeat wrote: »

Ok.. so f(x) = 4^x ; f'(x) = 4x.. similarly f(x) = e^x; f'(x) = ex = e^x ?

Not quite, but nearly. If f(x)=4^x, then f'(x) is a multiple of 4^x. It turns out that this multiple is ln(4), the natural log of 4.

In general, if f(x)=a^x, then f'(x) = (a^x)*Ln(a).

In the case of e^x, Ln(e) = 1.

MathsManiac

Just to clarify, I said that the derivative is proportional to the value of the function, (not to the value of x). So, in the case of 2^x, I meant that it's derivative is a multiple of 2^x, not that it was a multiple of x.

If y=a^x, then the dy/dx is a constant multiple of y.

sponsoredwalk

edit: Didn't see you got a response as I was busy drawing a picture ,
might be of use anyway...

"e" as in e¹ or e² is a constant, ℯˣ when the function is a function of x,
i.e. y = f(x) = ℯˣ means that the function changes as x changes &
there is a specific rate of change of ℯˣ with respect to x at different
times. I think you can think of ℯˣ as f(x) = (2.71828...)ˣ.

As for partial derivatives, well you have me confused, is it
f(x,y) = (1/x)y or f(x,y) = 1/(xy) ?
If it is f(x,y) = (1/x)y it's much clearer to write it as
f(x,y) = (y/x).
If it is f(x,y) = 1/(xy) when you give y a value, like 2, it's much clearer
to write it as
f(x,y) = 1/(2x)
In any case the derivative with respect to x means that you should
imagine a graph:



see how the rectangles get closer together, it signifies we're going
higher as we go to the center, like a mountain, & the straight
black line with the red covering it shows a person starting at the
bottom of the mountain walking straight up.
Here they are not moving in the y-direction at all so we hold it
constant. The red signifies the value of the derivative, it gets fatter to
signify the value gets bigger & the gradient is steeper.

So, if the rate of change just in the x-direction is

f(x,y) = 1/(2x) = (x⁻¹)/2 = ½(x⁻¹) then the derivative is???

Notice if we change the y-value up higher we'll get closer to the
center & the steepness will change more drastically as we get higher.

HellishHeat

Thank you both for your insights. I am enlightened.

MathsManiac

No problem.

Sponsoredwalk is correct to suggest that it's helpful to consider the partial derivative with respect to x visually, by considering moving parallel to the x-axis in the x-y plane.

For a 3D-plot of the function you're dealing with, click here:

http://www.wolframalpha.com/input/?i=Plot%5B1%2F%28x+y%29%2C+%7Bx%2C+-3.%2C+3.%7D%2C+%7By%2C+-3.%2C+3.%7D%5D

If you look at the 3D plot at the link above, the partial derivative wrt x tells you what's happening if you move along the surface shown, staying parallel to the x-axis.

You could also imagine taking a vertical slice through that 3D graph, where the slice is parallel to the x-axis, and you should be able to see that it cuts the surface in a shape that looks like the graph of 1/x (or a multiple of it). That's why the partial derivative of f(x,y) with respect to x is a "constant" multiple of 1/x, (with the "constant" being different for each value of y).