Differential Calculus (Differentiation) - SS3 Mathematics Past Questions and Answers - page 1

\(y = x^{3}\) i.e., \(f(x) = x^{3}\)

If \(x\) changes by \(\mathrm{\Delta}x\) then \(f(x + \mathrm{\Delta}x) = (x + \mathrm{\Delta}x)^{3}\)

\[\frac{dy}{dx} = \lim_{\mathrm{\Delta}x \rightarrow 0}\frac{f(x\ + \mathrm{\Delta}x) - f(x)}{\mathrm{\Delta}x}\]

\[\frac{dy}{dx} = \lim_{\mathrm{\Delta}x \rightarrow 0}\frac{(x + \mathrm{\Delta}x)^{3} - x^{3}}{\mathrm{\Delta}x}\]

\[\frac{dy}{dx} = \lim_{\mathrm{\Delta}x \rightarrow 0}\frac{(x + \mathrm{\Delta}x)^{3} - x^{3}}{\mathrm{\Delta}x}\]

\[\frac{dy}{dx} = \lim_{\mathrm{\Delta}x \rightarrow 0}\frac{x^{3} + 3x^{2}\mathrm{\Delta}x + 3x(\mathrm{\Delta}x)^{2} + (\mathrm{\Delta}x)^{3} - x^{3}}{\mathrm{\Delta}x}\]

\[\frac{dy}{dx} = \lim_{\mathrm{\Delta}x \rightarrow 0}\frac{3x^{2}\mathrm{\Delta}x + 3x(\mathrm{\Delta}x)^{2} + (\mathrm{\Delta}x)^{3}}{\mathrm{\Delta}x}\]

\[\frac{dy}{dx} = \lim_{\mathrm{\Delta}x \rightarrow 0}{3x^{2} + 3x\mathrm{\Delta}x + (\mathrm{\Delta}x)^{2}}\]

\(\mathrm{\Delta}x\) vanishes as \(\mathrm{\Delta}x \rightarrow 0\)

\(\therefore\frac{dy}{dx} = 3x^{2}\)

\[y = x^{5} + 4x^{3} - 5x + 4\]

\[\frac{dy}{dx} = \frac{d}{dx}(x^{5} + 4x^{3} - 5x + 4)\]

\[\frac{dy}{dx} = \frac{d}{dx}\left( x^{5} \right) + \frac{d}{dx}\left( 4x^{3} \right) - \frac{d}{dx}(5x) + \frac{d}{dx}(4)\]

\(\frac{dy}{dx} = 5x^{4} + 12x^{2} - 5 + 0 = 5x^{4} + 12x^{2} - 5\)

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

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

\[u = (2x + 1),\ \ \ \ \ \ \ v = (x^{2} + 2)\]

\[\frac{du}{dx} = 2,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dv}{dx} = 2x\]

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

\[\frac{dy}{dx} = 4x^{2} + 2x + 2x^{2} + 4\]

\(\frac{dy}{dx} = 6x^{2} + 2x + 4 = 2\lbrack 3x^{2} + x + 2\rbrack\)

 

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

\[u = 1,\ \ \ \ \ \ \ \ \ \ \ \ v = x\]

\[\frac{du}{dx} = 0,\ \ \ \ \ \ \ \ \ \frac{dv}{dx} = 1\]

\(\frac{dy}{dx} = \frac{x(0) - 1(1)}{x^{2}} = - \frac{1}{x^{2}}\)

 

Let \(u = 3x^{2} + 2\), \(\frac{du}{dx} = 6x\)

\(y = u^{2}\), \(\frac{dy}{du} = 2u\)

\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]

\[\frac{dy}{du} = 2u \times 6x = 2(3x^{2} + 2) \times 6x\]

\(= 12x(3x^{2} + 2)\)

 

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

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

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