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

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