Inverse Chain Rule

Introduction to inverse chain rule:

The chain rule is one of the most common rules used in the calculus for the calculation of the derivatives of the complex functions and the inverse chain rule is different that it is used for the calculation of the antiderivatives and the integrals of the functions. In the following article we will see in detail about the topic inverse chain rule.I like to share this Chain Rule Derivative with you all through my article.

More about the Topic Inverse Chain Rule:

As described before the chain rule is one of the most common rules for the calculation of the derivatives of the complex functions and the inverse chain rule is the inverse of the chain rule and is used for the calculation of the antiderivatives and the integrals of the functions. Here the Leibniz notation of the substitution is used. For the function described below,

`int_a^b f(g(t))g'(t) dt = int_{g(a)}^{g(b)} f(x)dx`

We substitute g(t) as x. And the differentiation of x with respect to t yields,

`x = g(t)`

`dx/dt = g(t)`

The above relation can be written as,

`dx = g'(t)dt`

This way of integration by the substitution is called as the inverse chain rule.

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Example Problems on Inverse Chain Rule:

1. Find the integrals of the function `int_0^2 x*cos(x^2+1) dx` using the inverse chain rule method.

Solution:

The given function is `int_0^2 x*cos(x^2+1) dx`

Here we substitute `x^2+1 = v`

And now differentiating v with respect to x we get,

dv/dx = 2x

dv = 2x dx

x dx = dv/2

Also the limits changes to 1 to 5.

So substituting the values in the function,

`int_0^2 x*cos(x^2+1) dx = int_0^2 cos(x^2+1) (x dx)`

= `int_0^2 x*cos(v) dv/2`

= `1/2 int_1^5 cos(v) dv`

= `1/2(sin(5)-sin(1))`

2. Find the integrals of the function `int_0^1 sqrt{1-x^2} dx` using the inverse chain rule method.

Solution:

The given function is `int_0^1 sqrt{1-x^2} dx`

Here we substitute `x = sin (v)`

And now differentiating v with respect to x we get,

dx/dv = cos (v)

dx = cos(v) dv

And also the limits changes to 0 to pi/2.

So substituting the values in the function,

`int_0^(pi/2) (sqrt{1-sin^2 v}) dx` = `int_0^(pi/2) sqrt{1-sin^2(v)} cos(v) dv`

= `int_0^(pi/2) cos(v) cos(v) dv`

= `int_0^(pi/2) cos^2(v) dv`

= `pi/4`