Differentiate, with respect to x, \(x^{3} + 2x\) from the first principle.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
\(y = x^{3} + 2x ... (1)\)
Let an increment in x = \(\Delta x\) and an increment in y = \(\Delta y\).
Then, \(y + \Delta y = (x + \Delta x)^{3} + 2(x + \Delta x)\)
\(y + \Delta y = x^{3} + 3x^{2} \Delta x + 3x (\Delta x)^{2} + (\Delta x)^{3} + 2x + 2 \Delta x ... (2)\)
\((2) - (1) : \Delta y = 3x^{2} \Delta x + 3x (\Delta x)^{2} + (\Delta x)^{3} + 2 \Delta x\)
\(\frac{\Delta y}{\Delta x} = 3x^{2} + 3x \Delta x + (\Delta x)^{2} + 2\)
\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_ {\Delta x \to 0} \frac{\Delta y}{\Delta x}\)
\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_{\Delta x \to 0} (3x^{2} + 3x \Delta x + (\Delta x)^{2} + 2 = 3x^{2} + 2\)
\(y = x^{3} + 2x ... (1)\)
Let an increment in x = \(\Delta x\) and an increment in y = \(\Delta y\).
Then, \(y + \Delta y = (x + \Delta x)^{3} + 2(x + \Delta x)\)
\(y + \Delta y = x^{3} + 3x^{2} \Delta x + 3x (\Delta x)^{2} + (\Delta x)^{3} + 2x + 2 \Delta x ... (2)\)
\((2) - (1) : \Delta y = 3x^{2} \Delta x + 3x (\Delta x)^{2} + (\Delta x)^{3} + 2 \Delta x\)
\(\frac{\Delta y}{\Delta x} = 3x^{2} + 3x \Delta x + (\Delta x)^{2} + 2\)
\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_ {\Delta x \to 0} \frac{\Delta y}{\Delta x}\)
\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_{\Delta x \to 0} (3x^{2} + 3x \Delta x + (\Delta x)^{2} + 2 = 3x^{2} + 2\)