Find from first principle the differential coef... - SS3 Mathematics Differential Calculus (Differentiation) Question
Find from first principle, the differential coefficient of \(y = x^{3}\)
\(\3x^{2}\)
\(\12x^{3}\)
\({\ x}^{3}\)
2x
\(y = x^{3}\) 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 \(\mathrm{\Delta}x \rightarrow 0\)
\(\therefore\frac{dy}{dx} = 3x^{2}\)
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