Question

Find the derivative of, (2x+3)³ using first principal

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \: f(x) = {(2x + 3)}^{3}$

So,

$\rm \: f(x + h) = {(2x + 2h + 3)}^{3}$

By using Definition of First Principal, we have

$\rm \: f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{f(x + h) - f(x)}{h}$

On substituting the values from above, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{ {(2x + 2h + 3)}^{3} - {(2x + 3)}^{3} }{h}$

can be further rewritten as

$\rm \: = \: 2\displaystyle\lim_{h \to 0}\rm \frac{ {(2x + 2h + 3)}^{3} - {(2x + 3)}^{3} }{2h}$

$\rm \: = \: 2\displaystyle\lim_{h \to 0}\rm \frac{ {(2x + 2h + 3)}^{3} - {(2x + 3)}^{3} }{2h + 2x - 2x}$

$\rm \: = \: 2\displaystyle\lim_{2x + 2h + 3 \to 2x + 3}\rm \frac{ {(2x + 2h + 3)}^{3} - {(2x + 3)}^{3} }{2h + 2x - 2x + 3 - 3}$

$\rm \: = \: 2\displaystyle\lim_{2x + 2h + 3 \to 2x + 3}\rm \frac{ {(2x + 2h + 3)}^{3} - {(2x + 3)}^{3} }{(2x + 2h + 3) - (2x + 3)}$

We know,

$\boxed{\tt{ \displaystyle\lim_{x \to a}\rm \: \frac{ {x}^{n} - {a}^{n} }{x - a} \: = \: {n \: a}^{n \: - \: 1} \: }}$

So, using this result, we get

$\rm \: = \: 2 \: \times \: 3 \: {(2x + 3)}^{3 - 1}$

$\rm \: = \: 6 \: {(2x + 3)}^{2}$

Hence,

$\rm\implies \:\boxed{\tt{ \dfrac{d}{dx} {(2x + 3)}^{3} = 6 {(2x + 3)}^{2} \: }}$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Question:-

Find the derivative of, (2x+3)³ using first principal.

Given:-

• The Equation (2x+3)³.

To Find:-

• The derivative of, (2x+3)³ using first principal.

Solution:-

$\sf \large \frac{d}{dx} = (2x + 3) {}^{3}$

$\sf \large = 3(2x + 3) {}^{2} . \frac{d}{dx} (2x + 3)$

$\sf \large = 3(2x + 3) {}^{2} .[ \frac{d}{dx} (2x) + \frac{d}{dx} (3)]$

$\sf \large = 3(2x + 3) {}^{2} [2 \frac{dx}{dx} + 0 ]$

$\sf \large = 3(2x + 3) {}^{2} (2)$

$\sf \large = 6(2x + 3) {}^{2}$

Answer:-

$\sf \large{ \boxed{ \sf \large \red{ \frac{d}{dx} (2x + 3) {}^{3} = 6(2x + 3) {}^{2} .}}}$

Hope you have satisfied. ⚘