Chain Rule (1 Viewer)

[tennille]

You may be given functions with different variables, and asked to find a specific derivative.

For example,

you are given x = t^2, and y = (3t^3) - 4

You are asked to find dy/dx.

dy/dx = dy/dt * dt/dx, since the dts cancel out.

dy/dt = 9t^2
dx/dt = 2t, so dt/dx = 1/2t

So dy/dx = 9t^2 * (1/2t)

dy/dx = 9t/2

This is a simple example. I hope that helps.

 

[Antwan23q]

chain rule?
isnt that like
y= uv
so the y'= u'v+v'u

 

[tennille]

chain rule?
isnt that like
y= uv
so the y'= u'v+v'u

That's the product rule when there are two functions multiplied by each other (like xsinx)- that's just a way to solve that.

 

[corbs18]

In maths study i keep coming across this chain rule and have no idea how to use it. can someone please explain it in easy terms and provide some examples please.

 

[acmilan]

Chain rule:

If y = [f(x)]ⁿ then y' = [f(x)]^n-1.f'(x)

Example:

y = [sinx + x]³

In this case, f(x) = sinx + x and f'(x) = cosx + 1

So y' = f(x)².f'(x) = [sinx + x]².(cosx+1)

 

[mattchan]

In simple turns for example ...

y = 3 (x - 1)^5

1: You drag the first down and minus one from the power

y = 3 x 5 ( x - 1 )^4

2: Then you differentiate whatever is inside the bracket

y = 3 x 5 ( x - 1)^4 x 1

3: Simplify

y = 15 (x -1)^4

Hope that helps

 

[goan_crazy]

Chain rule:
If y = [f(x)]ⁿ then y' = [f(x)]^n-1.f'(x)

I like that rule, the proper chain rule I don't use. I use that above rule. I think its called function of a function rule

 

[switchblade87]

I think its called function of a function rule

That is correct goan_crazy.
Simply put, its just the derivative of the function multiplied by the inside derivative of the bracket.

 

[haboozin]

You may be given functions with different variables, and asked to find a specific derivative.

For example,

you are given x = t^2, and y = (3t^3) - 4

You are asked to find dy/dx.

dy/dx = dy/dt * dt/dx, since the dts cancel out.

dy/dt = 9t^2
dx/dt = 2t, so dt/dx = 1/2t

So dy/dx = 9t^2 * (1/2t)

dy/dx = 9t/2

This is a simple example. I hope that helps.

jeez i dont think they would give you that...

only time i've ever seen a question like that was in 1994 4unit exam...

 

[rama_v]

Just remember though, that sometimes you have to use a combination of product rule and function of a function rule. For example

y = 3x(x+3)³

Sometimes people presume that the derivative is:

y' = 3x(3)(x+3)² * 1

This is incorrect. The correct way to do it is:

y' = u'v + uv' where u = 3x, v = (x+3)³

so:

y' = (3)(x+3)³ + (3x)(3(x+3)²*1)

y' = 3(x+3)³ + 9x(x+3)²

y' = 3(x+3)²((x+3) + 9x)

y' = 3(x+3)²(10x + 3)

 

[acmilan]

function of a function = chain rule. Its the same thing just different name:

y = [f(x)]ⁿ

Let u = f(x), du/dx = f'(x)

y = ⁿ
dy/dx = dy/du . du/dx = n^n-1.f'(x) = n[f(x)]^n-1.f'(x)

 

[tennille]

jeez i dont think they would give you that...

only time i've ever seen a question like that was in 1994 4unit exam...

The example I gave is simple (maybe it looks hard because its on the computer). And I never did 4 unit maths to see the relation to it.

You'll see similar examples in 2 unit.

In 2 unit, I learnt chain rules that involved 3 derivatives, as well as 2.

 

[icycloud]

jeez i dont think they would give you that...

only time i've ever seen a question like that was in 1994 4unit exam...

The chain rule in the form dy/dx = dy/du * du/dx is in the 2U syllabus.

 

[PoP 'n' Fresh]

more commonly seen in 3u