Rules Of Differentiation - SS3 Mathematics Lesson Note

1. SUM RULE: If \(y = u + v\), where \(u\) and \(v\) are functions of \(x\) then

\[\frac{dy}{dx} = \frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx}\]

1. DIFFERENCE RULE: If \(y = u - v\), where \(u\) and \(v\) are functions of \(x\) then

\[\frac{dy}{dx} = \frac{d}{dx}(u - v) = \frac{du}{dx} - \frac{dv}{dx}\]

1. PRODUCT RULE: If \(y = uv\), where \(u\) and \(v\) are functions of \(x\) then

\[\frac{dy}{dx} = \frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\]

1. QUOTIENT RULE: If \(y = \frac{u}{v}\), where \(u\) and \(v\) are functions of \(x\) then

\[\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}\]

1. CHAIN RULE: Suppose \(y\ \)is a function of a variable, \(u\) and \(u\) is a function of \(x\), then \(y\) is a function of a function of \(x\). Symbolically, if \(y = f(u)\) and \(u = g(x)\), then \(y = f(g(x))\). The derivative of \(y\) is \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)

Example 3 Find the differential coefficients of the following function: (a) \(y = 6x^{3} - \cos{x + 5\sin{x - 2e^{x}}}\) (b) \(y = x^{2}\sin x\) (c) \(y = \frac{\cos x}{1 + \sin x}\) (d) \(y = \sin{(3x + 2)} + e^{2x}\)

Solution

1. \(y = 6x^{3} - \cos{x + 5\sin{x - 2e^{x}}}\)

\[\frac{dy}{dx} = \frac{d}{dx}(6x^{3} - \cos{x + 5\sin{x - 2e^{x}}})\]

\[\frac{dy}{dx} = \frac{d}{dx}\left( 6x^{3} \right) - \frac{d}{dx}\left( \cos x \right) + \frac{d}{dx}\left( 5\sin x \right) - \frac{d}{dx}(2e^{x})\]

\[\frac{dy}{dx} = 18x^{2} - \left( - \sin x \right) + 5\cos x - 2e^{x}\]

\[\frac{dy}{dx} = 18x^{2} + \sin x + 5\cos x - 2e^{x}\]

1. \(y = x^{2}\sin x\)

\[\frac{dy}{dx} = x^{2}.\frac{d\left( \sin x \right)}{dx} + \sin x\frac{d\left( x^{2} \right)}{dx}\]

\[\frac{dy}{dx} = x^{2}(\cos{x)} + \sin x(2x)\]

\[\frac{dy}{dx} = x^{2}\cos x + 2x\sin x\]

1. \(y = \frac{\cos x}{1 + \sin x}\)

\[\frac{dy}{dx} = \frac{d}{dx}\left( \frac{\cos x}{1 + \sin x} \right)\]

\[\frac{dy}{dx} = \frac{1 + \sin x\left( - \sin x \right) - \cos x(\cos x)}{\left( 1 + \sin x \right)^{2}}\]

\[\frac{dy}{dx} = \frac{- \sin x - \sin^{2}x - \cos^{2}x}{\left( 1 + \sin x \right)^{2}}\]

\[\frac{dy}{dx} = \frac{- \sin x - (\sin^{2}x + \cos^{2}x)}{\left( 1 + \sin x \right)^{2}}\]

\[\frac{dy}{dx} = \frac{- \sin x - (1)}{\left( 1 + \sin x \right)^{2}}\]

\[\frac{dy}{dx} = \frac{- 1 - \sin x}{\left( 1 + \sin x \right)^{2}}\]

\[\frac{dy}{dx} = \frac{- (1 + \sin x)}{\left( 1 + \sin x \right)^{2}}\]

\[\frac{dy}{dx} = \frac{- 1}{1 + \sin x}\]

1. \(y = \sin{(3x + 2)} + e^{2x}\)

\[y = y_{1} + y_{2}\]

For the first term, \(\sin(3x + 2)\): \(y_{1} = \sin(3x + 2)\), \(u_{1} = 3x + 2\)

\[y_{1} = \sin u_{1}\]

For the second term, \(e^{2x}\): \(y_{2} = e^{2x}\), \(u_{2} = 2x\)

\[y_{2} = e^{u_{2}}\]

\[\frac{dy_{1}}{dx} = \frac{dy_{1}}{du_{1}}.\frac{du_{1}}{dx}\]

\[\frac{dy_{1}}{dx} = \cos u_{1} \times 3 = 3\cos u_{1} = 3\cos{(3x + 2)}\]

\[\frac{dy_{2}}{dx} = \frac{dy_{2}}{du_{2}}.\frac{du_{2}}{dx}\]

\[\frac{dy_{2}}{dx} = e^{u_{2}} \times 2 = 2e^{u_{2}} = 2e^{2x}\]

\[y = y_{1} + y_{2}\]

\[\frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{1}}{dx} = 3\cos{(3x + 2)} + 2e^{2x}\]