Here you will learn what is chain rule in differentiation with examples.
Letโs begin โ
Chain Rule in Differentiation
If f(x) and g(x) are differentiable functions, then fog is also differentiable and
(fog)'(x) = f'(x) = f'(g(x)) g'(x)
or, \(d\over dx\) {(fog) (x)} = \(d\over d g(x)\) {(fog) (x)} \(d\over dx\) (g(x)).
Remark 1ย : The above rule can also be restated as follows :
If z = f(y) and y = g(x), then \(dz\over dx\) = \(dz\over dy\).\(dy\over dx\)
Derivative of z with respect to x = (Derivative of z with respect to y) \(\times\) (Derivative of y with respect to x)
Remark 2 : This chain rule can be extended further.
Derivative of z with respect to x = (Derivative of z with respect to u) \(\times\) (Derivative of u with respect to v) \(\times\) (Derivatve of v with respect to x)
Example 1 : Differentiate \(sin(x^2 + 1)\) with respect to x.
Solution : Let y = \(sin(x^2 + 1)\). Putting u = \(x^2 + 1\) , we get
y = sin u and u = \(x^2 + 1\)
\(\therefore\)ย \(dy\over du\) = cosu and \(du\over dx\) = 2x
Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)
\(\implies\) \(dy\over dx\) = (cos u)2x = 2x \(cos (x^2 + 1)\)
Hence, \(d\over dx\) [\(sin(x^2+1)\)] = 2x \(cos (x^2 + 1)\)
Example 2 : Differentiate \(e^{sinx}\) with respect to x.
Solution : Let y = \(e^{sinx}\). Putting u = sinx , we get
y = \(e^u\) and u = sinx
\(\therefore\)ย \(dy\over du\) = \(e^u\) and \(du\over dx\) = cosx
Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)
\(\implies\) \(dy\over dx\) = \(e^u\)cosx = \(e^{sinx}\)cosx
Hence, \(d\over dx\) [\(e^{sinx}\)] = \(e^{sinx}\)cosx
Example 3 : Differentiate log sinx with respect to x.
Solution : Let y = log u. Putting u = sinx , we get
y = log u and u = sinx
\(\therefore\)ย \(dy\over du\) = \(1\over u\) and \(du\over dx\) = cosx
Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)
\(\implies\) \(dy\over dx\) = \(1\over u\)cosx = \(1\over sinx\)cosx = cotx
Hence, \(d\over dx\) [log sinx] = cotx