The chain rule
The chain rule is used to differentiate composite functions. It is written as:
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \times \frac{{du}}{{dx}}\)
Example (extension)
Differentiate \(y = {(2x + 4)^3}\)
Solution
Using the chain rule, we can rewrite this as:
\(y = {(u)^3}\) where \(u = 2x + 4\)
We can then differentiate each of these separate expressions:
\(\frac{{dy}}{{du}} = 3{(u)^2}\) and \(\frac{{du}}{{dx}} = 2\)
Therefore:
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \times \frac{{du}}{{dx}}\)
\(\frac{{dy}}{{dx}} = 3{(u)^2} \times 2\)
\(\frac{{dy}}{{dx}} = 6{(u)^2}\)
\(\frac{{dy}}{{dx}} = 6{(2x + 4)^2}\)
Question
Extension
Differentiate \(y = {(2{x^2} + 3x + 4)^{\frac{1}{2}}}\)
Using the chain rule, we can rewrite this as:
\(y = {(u)^{\frac{1}{2}}}\) where \(u = 2{x^2} + 3x + 4\)
\(\frac{{dy}}{{du}} = \frac{1}{2}{(u)^{ - \frac{1}{2}}}\)
\(\frac{{du}}{{dx}} = 4x + 3\)
Therefore:
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \times \frac{{du}}{{dx}}\)
\(\frac{{dy}}{{dx}} = \frac{1}{2}{(u)^{ - \frac{1}{2}}} \times (4x + 3)\)
\(\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt u }} \times (4x + 3)\)
\(\frac{{dy}}{{dx}} = \frac{{4x + 3}}{{2\sqrt {2{x^2} + 3x + 4} }}\)
Question
Extension
Differentiate \(y = {(1 + \sin x)^3}\)
Using the chain rule, we can rewrite this as:
\(y = {(u)^3}\) where \(u=1+sinx\)
\(\frac{{dy}}{{du}} = 3{(u)^2}\)
\(\frac{{du}}{{dx}} = \cos x\)
Therefore:
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \times \frac{{du}}{{dx}}\)
\(\frac{{dy}}{{dx}} = 3{(u)^2} \times \cos x\)
\(\frac{{dy}}{{dx}} = 3\cos x{(u)^2}\)
\(\frac{{dy}}{{dx}} = 3\cos x{(1 + \sin x)^2}\)
Question
Extension
Differentiate \(y = \sin (3{x^2} + 4)\)
Using the chain rule, we can rewrite this as:
\(y = \sin u\) where \(u = 3{x^2} + 4\)
\(\frac{{dy}}{{du}} = \cos u\)
\(\frac{{du}}{{dx}} = 6x\)
Therefore:
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \times \frac{{du}}{{dx}}\)
\(\frac{{dy}}{{dx}} = \cos u \times 6x\)
\(\frac{{dy}}{{dx}} = 6x\cos u\)
\(\frac{{dy}}{{dx}} = 6x\cos (3{x^2} + 4)\)
Question
Extension
Differentiate \(y = {(\sin 2x)^2}\)
Using the chain rule, we can rewrite this as:
\(y = {(u)^2}\) where \(u = \sin 2x\)
\(\frac{dy}{du}=2u\)
\(\frac{du}{dx}=2cos2x\)
Therefore:
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \times \frac{{du}}{{dx}}\)
\(\frac{{dy}}{{dx}} = 4u\cos 2x\)
\(\frac{{du}}{{dx}} = 4\sin 2x\cos 2x\)
