When a function is a composite of two or more functions then to find the derivative of that function, we use chain rule. Let us understand the proof of chain-rule:
Let y = f(u)
And z= g(x)
where y and z both are differentiable functions.
Then y= f(g(x)) is a function of x
delta y = f(u+delta u ) – f(u)
And delta z = g(x+delta x) – g(x)
Therefore delta y/ delta z = f(u+delta u ) – f(u)/delta t
And delta z/ delta x = g(x+ delta x) – g(x)/delta x
We first suppose that for sufficiently small values of delta x , delta z not equal to 0. This will be the case if dz/dx not equal to 0. Then, g is differentiable, it is continuous and hence when delta x ->0, x+delta x->x and g(x+delta x)->g(x).
Therefore z+delta z->z and delta z->0. Since delta z not equal to 0 as delta x->0, we may write
delta y/delta x= delta y/delta u . delta z/delta x
We can see that f and g, both are continuous functions as they are differentiable. Thus, delta u->0 when delta x->0 and delta y->0 when delta u->0.
Therefore, limit delta x->0 delta y/delta x = limit delta z->0 delta y/delta z multiply limit delta x->0 delta t/delta x
Hence dy/dx = dy/dz. dz/dx
For example, find derivative of sin(x^2)
Solution: Here we can clearly see that the function is function of a function.
Suppose y = sin(x^2) = sin z
Where z = x^2
Therefore dy/dz = cos z and dz/dx = 2x
Hence by this rule
Dy/dx = dy/dz. dz/dx
= cosz . 2x [ substitute the values of dy/dz and dz/dx]
= 2xcos(x^2) [ substitute the value of z]
Derivatives chain rule: Sometimes it is also known as chain-rule derivatives. It means to find derivative of a function using this rule.
Reverse Chain Rule:
It is used for integration or say anti derivative of functions. It is equivalent to integration by parts.
The Chain Rule Multivariable:
It means many variables are included in this form. Suppose that y=f (a, b), where a’ and b themselves depend on one or more variables. This Rule gives us the liberty to differentiate y with respect to any of the variables (a, b) involved:
Let a=a (t) and b=b (t) be differentiable at t and let y=f (a, b) is differentiable at the point (a (t), b (t)). Then y=f (a (t), b (t)) is differentiable at t and
Dy/dt = dy/ da . da/dt + dy/ db . db/dt
Let us understand an example of the multivariate chain-rule:
For example: Solve it using the multivariate chain-rule: Let y = a^2b-b^2 where a and b are parameterized as a=t^2 and b=2t
Then Dy/dt = dy/ da . da/dt + dy/ db . db/dt
=2ab.2t +(a^2-2b)(2)
= (2t^2. 2t).2t+((t^2)^2 – 2(2t))(2)
= 8t^4 + 2t^4 – 8t
= 10t^4 -8t
