First slide

Diffrentiation

Question

$If y = x^{3} + 2 x^{2} + 7 x + 8 then \frac{d y}{d x} will be$

Easy

A

$3 x^{2} + 2 x + 15$
B

$3 x^{2} + 4 x + 7$
C

$x^{3} + 2 x^{2} + 15$
D

$x^{3} + 4 x + 7$

Solution

$\begin{array}{l} y = x^{3} + 2 x^{2} + 7 x + 8 \\ differentiate on both sides w.r.t ' x^{'} \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3} + 2 x^{2} + 7 x + 8) \\ = 3 x^{2} + 2 (2 a) + 7 (1) + 0 [∵ \frac{d c}{d x} = 0, \frac{d x}{d x} = 1, \frac{d x^{n}}{d x} = n x^{n - 1}] \\ = 3 x^{2} + 4 x + 7 \end{array}$

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