Question

Can someone explain the chain rule?

Answer 1

See below.

Explanation:

First, I assume you mean the chain rule.

Applied when we want to find the derivative of a function of a function of the variable.

Let's assume that we have a function `F(x)` defined as
`F(x) = f(g(x))`

where `f` and `g` are both functions of `x`

Then the chain rule states that:

`F'(x) = f'(g(x)) * g'(x)`

As an example, let's consider `y= e^(2x)` and we want to find the first derivative, `dy/dx`

Using the nomenclature above:

`F(x)=y, f(x) = e^x and g(x) = 2x`

Then note that: `F'(x) = dy/dx, f'(x) = e^x and g'(x) =2`

Hence, applying the chain rule:

`dy/dx = e^(2x)*2 = 2e^(2x)`

I hope this helps.

[After a little practice you'll find you will be able to apply this rule in much more complicated cases without needing to go back to the definition.]

Answer 2

Please see below.

Explanation:

We may have done different kinds of differentiation when a function is say `f(x)=x^3+x^2+2` or `g(x)=tanx` or `h(x)=sqrtx` or `t(x)=lnx` or `u(x)=e^x` or similar functions.

But how does one differentiate when we have a combination of functions such as `sqrt(x^3+x^2+2)` or `ln(tanx)` or `e^sqrtx`.

Observe that we can write them as `h(f(x))` or `t(g(x))` or `u(h(x))`. Hence such functions are called function of a function. Sometimes it could be more complicated i.e. function of a function of a function. For example, let us say `ln[tan(sqrt(x^3+x^2+2))]` which is just `t(g(f(x)))`.

How do we deal with such function of a function of another function?

In order to differentiate such complicated functions, we need to use Chain Rule, according to

when we have `y=y(u(x))` then `(dy)/(dx)=(dy)/(du)xx(du)/(dx)`

For example, for `y=sqrt(x^3+x^2+2)`, where `f(x)=sqrtx` and `g(x)=x^3+x^2+2`, then `y=f(g(x))` and

`(dy)/(dx)=(df)/(dg)xx(dg)/(dx)`

As `f=sqrtg`, we have `(df)/(dg)=1/(2sqrtg)` and as `g=x^3+x^2+2`, `(dg)/(dx)=3x^2+2x`

and `(dy)/(dx)=1/(2sqrtg)xx(3x^3+2x)=(3x^3+2x)/(2sqrt(x^3+x^2+2)` as `g=x^3+x^2+2`

Similarly when we have `y=f(g(h(x)))`, we have

`(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)`

For example for `y=ln[tan(sqrt(x^3+x^2+2))]`

`(dy)/(dx)=1/tan(sqrt(x^3+x^2+2))xxsec^2(sqrt(x^3+x^2+2))xx1/(2sqrt(x^3+x^2+2)xx(3x^2+2x)`