The differential is written as dy/dx and has three distinct and yet simultaneous meanings:
dy/dx [“dy by dx”] can just as easily refer to the function that the differential represents;
dy / dx [“dy over dx”] can be a fraction, a division of dy by dx; and we have the
d/dx (y) [“d by dx of y"] which in long form is ‘the differential with respect to x of the function y’ or, slightly shorter
‘d by dx of y’. We use all three forms interchangeably.
In general d/dx (axⁿ) = anxn-1 which is sufficient to cope with polynomials. You should prove that the differential of f + g, where f & g are both functions in x, is the sum of their differentials.
The differential dy/dx = f’(x) can be separated to give dy = f’(x) dx. This shows the relationship between the small part of y, dy, and the corresponding and very small dx. All these small parts can be added up, a process called summation, to form the integral, written ∫dy = ∫ f’(x) dx. Pretty obviously, adding up all the little bits of y and you really should get y, so ∫dy = y and ∫ f’(x) dx is the integral to work out.
If you look back at differentiation, d/dx (axⁿ) = anxn-1 so therefore ∫ anxn-1 dx = axⁿ and, this might sensibly be written as ∫ anxn-1 dx = axⁿ / n+1 . A constant can be added at the end because if you try differentiating your answer, you will rediscover that the differential of a constant is zero. If you think of the differential as being the gradient function, then y = constant has a gradient of zero, so no differential. The rule you tell yourself for integration is “add one to the index and divide by the new index”.
While you can differentiate any function (eventually) there are many functions that you cannot integrate. However, before we get caught up in that, let us just spend a moment looking at what we can do with the notation:
∫ dy says add up all the little bits of y. Look at a graph: what does this mean? Not a lot…
How about ∫ x dy ? What might ∫ x dy mean? How about it might mean a little strip across the page of length x and height dy? Then adding this up would give an area, ∫ x dy. In the same way, so would ∫ y dx, but the strips would be vertical ones. Many students remain stuck with this as their only concept of integration. What might ∫ πy² dx mean? Would π y² mean a disk of radius y and π y² dx is therefore a disk of thickness dx? So adding them up, namely doing ∫ π y² dx, must mean starting with a function y and spinning it around the x-axis to make a solid; ∫ π y² dx calculates the volume of that solid. I prefer to write π ∫ y² dx, because pi is a constant and doesn’t affect the integration process. Think of the ∫ and dx as brackets, delimiting the thing to be integrated. the last x is important, as it explains what variable we are using.
Chain Rule: for a function in y (not x) then df/dx = d/dx f(y) = df/dy . dy/dx
You can see the two dy terms cancelling. The trick is that by inserting them you can succeed in doing the differentiation.
For example d/dx (3x – 6)³ could be treated as f(y) = y³ and y = 3x-6,
So df/dy = 3y² and dy/dx = 3 so d/dx (3x – 6)³ = 3 (3x – 6)² . 3
Product Rule: for two functions u(x) and v(x) d/dx (u.v) = u. dv/dx + v . du/dx
For a proof, go back to the x form: y = u.v and both u and v are functions of x. If ∂x is a small increase in x then ∂u, ∂v and ∂y are corresponding small increases in u, v and y.
y + 𝜕y = (u + 𝜕u) (v + 𝜕v) = uv + u.∂v + v.∂u + ∂u.∂v we know y = uv
Therefore ∂y = ∂u.v + ∂v.u + ∂u.∂v and note that ∂u∂v is the square of ‘small’
Divide all by ∂x, so ∂y/∂x = u.∂v/x + v.∂u/x + ∂u. ∂v/∂x
Let ∂x tend to zero: so all ∂ become d and the last term disappears as u becomes zero.
Thus ∂y/∂x = u ∂v/∂x + v ∂u/∂x + ∂u. ∂v/∂x becomes dy/dx = u dv/dx + v du/dx
To create the Quotient rule, put v-1 in place of v and apply the chain rule. Voila !
Note that dx/dy is the reciprocal of dy/dx. A direct use of this follows when y = ln x, so x = ey and dx/dy = ey = x because this gives that dy/dx = d/dx (ln x) = 1/x
See also the pages on Exponentials and Integration
General revision and editing 201605 following home machine crash.
Area & Perimeter (application)
