{"id":5127,"date":"2021-09-08T18:29:08","date_gmt":"2021-09-08T12:59:08","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5127"},"modified":"2021-11-22T22:07:12","modified_gmt":"2021-11-22T16:37:12","slug":"chain-rule-in-differentiation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/chain-rule-in-differentiation\/","title":{"rendered":"Chain Rule in Differentiation with Examples"},"content":{"rendered":"
Here you will learn what is chain rule in differentiation with examples.<\/p>\n
Let’s begin –<\/p>\n
If f(x) and g(x) are differentiable functions, then fog is also differentiable and<\/p>\n
\n(fog)'(x) = f'(x) = f'(g(x)) g'(x)<\/p>\n
or, \\(d\\over dx\\) {(fog) (x)} = \\(d\\over d g(x)\\) {(fog) (x)} \\(d\\over dx\\) (g(x)).<\/p>\n<\/blockquote>\n
Remark 1 <\/strong>: The above rule can also be restated as follows :<\/p>\n
\nIf z = f(y) and y = g(x), then \\(dz\\over dx\\) = \\(dz\\over dy\\).\\(dy\\over dx\\)<\/p>\n
Derivative of z with respect to x = (Derivative of z with respect to y) \\(\\times\\) (Derivative of y with respect to x)<\/p>\n<\/blockquote>\n
Remark 2<\/strong> : This chain rule can be extended further.<\/p>\n
\nDerivative 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)<\/p>\n<\/blockquote>\n
Example 1<\/span><\/strong> : Differentiate \\(sin(x^2 + 1)\\) with respect to x.<\/p>\n
Solution<\/span><\/strong> : Let y = \\(sin(x^2 + 1)\\). Putting u = \\(x^2 + 1\\) , we get<\/p>\n
y = sin u and u = \\(x^2 + 1\\)<\/p>\n
\\(\\therefore\\) \\(dy\\over du\\) = cosu and \\(du\\over dx\\) = 2x<\/p>\n
Now, \\(dy\\over dx\\) = \\(dy\\over du\\) \\(\\times\\) \\(du\\over dx\\)<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = (cos u)2x = 2x \\(cos (x^2 + 1)\\)<\/p>\n
Hence, \\(d\\over dx\\) [\\(sin(x^2+1)\\)] = 2x \\(cos (x^2 + 1)\\)<\/p>\n
Example 2<\/span><\/strong> : Differentiate \\(e^{sinx}\\) with respect to x.<\/p>\n
Solution<\/span><\/strong> : Let y = \\(e^{sinx}\\). Putting u = sinx , we get<\/p>\n
y = \\(e^u\\) and u = sinx<\/p>\n
\\(\\therefore\\) \\(dy\\over du\\) = \\(e^u\\) and \\(du\\over dx\\) = cosx<\/p>\n
Now, \\(dy\\over dx\\) = \\(dy\\over du\\) \\(\\times\\) \\(du\\over dx\\)<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = \\(e^u\\)cosx = \\(e^{sinx}\\)cosx<\/p>\n
Hence, \\(d\\over dx\\) [\\(e^{sinx}\\)] = \\(e^{sinx}\\)cosx<\/p>\n
Example 3<\/span><\/strong> : Differentiate log sinx with respect to x.<\/p>\n
Solution<\/span><\/strong> : Let y = log u. Putting u = sinx , we get<\/p>\n
y = log u and u = sinx<\/p>\n
\\(\\therefore\\) \\(dy\\over du\\) = \\(1\\over u\\) and \\(du\\over dx\\) = cosx<\/p>\n
Now, \\(dy\\over dx\\) = \\(dy\\over du\\) \\(\\times\\) \\(du\\over dx\\)<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = \\(1\\over u\\)cosx = \\(1\\over sinx\\)cosx = cotx<\/p>\n
Hence, \\(d\\over dx\\) [log sinx] = cotx<\/p>\n\n\n