Find the differential coefficient of the follow... - SS3 Mathematics Differential Calculus (Differentiation) Question

(a). \(y = 3x\ \left( x^{2} + 1 \right)\sin x\)

Resolving the first two functions, \(3x\) and \((x^{2} + 1)\) by product rule, then the third \(\sin x\)

Let \(u = 3x\) and \(v = (x^{2} + 1)\)

\[u\frac{dv}{dx} + v\frac{du}{dx}\]

\[= 3x\ (2x) + (x^{2} + 1)3\]

\[= 6x^{2} + 3x^{2} + 3\]

\[= 9x^{2} + 3\]

Resolving \(u = 9x^{2} + 3\)and \(v = \sin x\)

\[u\frac{dv}{dx} + v\frac{du}{dx}\]

\[= 9x^{2} + 3\left( \cos x \right) + \sin x(18x)\]

\[\frac{dy}{dx} = 9x^{2} + 3\left( \cos x \right) + 18x\sin x\]

\[\frac{dy}{dx} = 3\lbrack 3x^{2} + \cos x + 9x\sin x\rbrack\]

 

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

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

\(u = 1 + \tan x\), \(\frac{du}{dx} = \sec^{2}x\)

\(v = \sin x\), \(\frac{dv}{dx} = \cos x\)

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

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

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

 

(c). \(y = \log_{e}{(1 + x)}\)

Let \(u = 1 + x\), \(\frac{du}{dx} = 1\)

\(y = \log_{e}u\) \(\frac{dy}{du} = \frac{1}{u}\)

\(\frac{dy}{dx} = \frac{dy}{du}.\frac{du}{dx} = \frac{1}{u}.(1) = \frac{1}{u} = \frac{1}{1 + x}\)