Question

find the derivative of x³+4x²+3x+2 with respect to x​

Answer 1

To find :

The derivative of the function :-

$\bf{x^{3} + 4x^{2} + 3x + 2}$

Solution :

We know the formula for , Derivative by first principle i.e,

$\boxed{\bf{f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}}}$

Now using the above equation and substituting the values in it, we get :

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{(x + h)^{3} + 4(x + h)^{2} + 3(x + h) + 2 - (x^{3} + 4x^{2} + 3x + 2)}{h}}$

$\boxed{\begin{minipage}{7 cm}Using the identities :- \\ \\ :\implies \bf{(a + b)^{3} = a^{3} + b^{3} + 3a^{2}b+ 3ab^{2}} \\ \\ :\implies \bf{(a + b)^{2} = a^{2} + b^{2} + 2ab} \\ \\ we get :\end{minipage}}$

$:\implies \bf{f'(x) = \lim_{h \to 0}}$ ((x³ + h³ + 3x²h + 3xh²) + 4(x² + 2xh + h²) + 3(x + h) + 2 - (x³ + 4x² + 3x + 2)/h

$:\implies \bf{f'(x) = \lim_{h \to 0}}$ (x³ + h³ + 3x²h + 3xh² + 4x² + 8xh + h² + 3x + 3h + 2 - x³ - 4x² - 3x - 2)/h

$:\implies \bf{f'(x) = \lim_{h \to 0}}$ ((x³ - x³) + h³ + 3x²h + 3xh² + (4x² - 4x²) + 8xh + h² + (3x - 3x) + 3h + (2 - 2))/h

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{h^{3} + 3x^{2}h + 3xh^{2} + 8xh + 4h^{2} + 3h}{h}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{h^{3}}{h} + \dfrac{3x^{2}h}{h} + \dfrac{3xh^{2}}{h} + \dfrac{8xh}{h} + \dfrac{4h^{2}}{h} + \dfrac{3h}{h}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} h^{2} + \dfrac{3x^{2}\not{h}}{\not{h}} + 3xh + \dfrac{8x\not{h}}{\not{h}} + 4h + \dfrac{3\not{h}}{\not{h}}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} h^{2} + 3x^{2} + 3x + 8 + 4h + 3}$

$:\implies \bf{f'(x) = 0^{2} + 3x^{2} + 3xh + 8x + 4(0) + 3}$

$:\implies \bf{f'(x) = 0 + 3x^{2} + 3x(0) + 8x + 0 + 3}$

$:\implies \bf{f'(x) = 3x^{2} + 0 + 8x + 3}$

$:\implies \bf{f'(x) = 3x^{2} + 8x + 3}$

$\boxed{\therefore \bf{f'(x) = 3x^{2} + 8x + 3}}$

Hence the derivative of

$\bf{x^{3} + 4x^{2} + 3x + 2}$ is $\bf{3x^{2} + 8x + 3}$