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Derivative from First Principle

Find the deri...

Question

Find the derivative by the first principle of $f (x) = x^{3} + 4 x$

A

3 x^{2} + 4

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B

x \sqrt{x}

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C

\frac{1}{2 x + 3}

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D

\frac{x - 1}{2 x + 7}

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Solution

The correct option is A $3 x^{2} + 4$
$f (x) = x^{3} + 4 x$

$f^{'} (x) = l i m_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$= l i m_{h \to 0} \frac{(x + h)^{3} + 4 (x + h) - x^{3} - 4 x}{h} = l i m_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} + 4 x + 4 h - x^{3} - 4 x}{h}$

$= l i m_{h \to 0} \frac{(x + h)^{3} + 4 (x + h) - x^{3} - 4 x}{h} = l i m_{h \to 0} \frac{3 x^{2} h + 3 x h^{2} + h^{3} + 4 h}{h} = l i m_{h \to 0} \frac{h (3 x^{2} + 4 + h^{2} + 3 x h)}{h} = 3 x^{2} + 4$

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Similar questions

Q. Question 14
By Remainder theorem, find the remainder when p(x) is divided by g(x).

(i)

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

f (x) = {\begin{matrix} x^{2} - 4 x + 3, & x < 2 x - 4, & x \geq 2 \end{matrix}

g (x) = {\begin{matrix} x - 3, & x < 3 x^{2} + 2 x + 2, & x \geq 3 \end{matrix}

\frac{f}{g} \cdot (1)

Q. Question 14
By Remainder theorem, find the remainder when p(x) is divided by g(x).

(i)

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

Q. Find the derivative by first principle
(1)

3 x^{2} + 4

\frac{1}{2 x + 3}

Q. A polynomial

p (x)

g (x)

, the obtained quotient

q (x)

and the remainder

r (x)

are given in the table. Find

p (x)

Sl.	$p (x)$	$g (x)$	$q (x)$	$r (x)$
i	?	$x - 2$	$x^{2} - x + 1$	$4$
ii	?	$x + 3$	$2 x^{2} + x + 5$	$3 x + 1$
iii	?	$2 x + 1$	$x^{3} + 3 x^{2} - x + 1$	$0$
iv	?	$x - 1$	$x^{3} - x^{2} - x - 1$	$2 x - 4$
v	?	$x^{2} + 2 x + 1$	$x^{4} - 2 x^{2} + 5 x - 7$	$4 x + 12$

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