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HOW TO FIND THE SECOND DERIVATIVE ON CALCULUS USING DERIVATIVE FORMULA | CALCULUS |

Posted by : Allan_Dell on Thursday, March 30, 2023 | 10:26 PM

THE SECOND DERIVATIVE | BY THE DERIVATIVE FORMULA

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Differential calculus is a branch of calculus that focuses on studying the rates at which things change. It involves finding the derivative of a function, which represents the instantaneous rate of change at any given point. Differential calculus is used extensively in fields such as physics, engineering, economics, and finance to model and analyze real-world phenomena, including motion, optimization, and growth.

Well, this method is different from our previous one, the increment method, and by solving only the first derivative. In this section, we will be dealing with the second derivative. The second derivative is like deriving the previous derivation. It is similar to saying that we derive again after the first derivation. We will be using the formula above because nothing will be changed here. Let's dive in!

Given the equations below, find the second derivative of each.

$1.)\ x^2$

$x^2,\ @{\color{Red} }\ n=2$

$\frac{d}{dx}[x^{\color{DarkRed} 2}]={\color{DarkRed} 2}x^{2{\color{DarkRed} }-1}$

$\frac{d}{dx}[x^{\color{DarkRed} 2}]={\color{DarkRed} 2}x^1=\underline{{\color{DarkRed} 2}x}$ , first derivative

Now, solving the second derivative

$\frac{d}{dx}[2x]=2x^{1-1}=2x^0=2(1)=\underline{2}$ , second derivative (answer)

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$2.)\ 2x^3$

$2x^3\ @\ {\color{DarkRed} n=3}$

$\frac{d}{dx}[2x^{\color{DarkRed} 3}]={\color{DarkRed} 3}(2x)^{{\color{DarkRed} 3}-1}$

$\frac{d}{dx}[2x^{\color{DarkRed} 3}]=6x^2$ , first derivative

Now, solving the second derivative

$\frac{d}{dx}[6x^{\color{DarkRed} 2}]={\color{DarkRed} 2}(6x)^{{\color{DarkRed} 2}-1}$

$\frac{d}{dx}[6x^{\color{DarkRed} 2}]={\color{DarkRed} 2}(6x)^{{\color{DarkRed} 2}-1}=12x^1=\underline{12x}$ , second derivative (answer)

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$3.)\ \frac{1}{2}x^{2}$

$\frac{1}{2}x^{2}\ @\ {\color{DarkRed} n=2}$

$\frac{d}{dx}[\frac{1}{2}x^{\color{DarkRed} 2}]={\color{DarkRed} 2}\frac{1}{2}x^{{\color{DarkRed} 2}-1}=x^1=x$ , first derivative*

Now, solving the second derivative

$\frac{d}{dx}[x^{\color{DarkRed} 1}]={\color{DarkRed} 1}*x^{{\color{DarkRed} 1}-1}={\color{DarkRed} 1}*x^0={\color{DarkRed} 1}*1=\underline{1}$ , second derivative (answer)

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$4.)\ 2x^\frac{1}{2}$

$2x^\frac{1}{2}\ @\ {\color{DarkRed} n=\frac{1}{2}}$

$\frac{d}{dx}[2x^{\color{DarkRed} \frac{1}{2}}]={\color{DarkRed} \frac{1}{2}}x^{{\color{DarkRed} \frac{1}{2}}-1}$

$\frac{d}{dx}[2x^{\color{DarkRed} \frac{1}{2}}]={\color{DarkRed} \frac{1}{2}}x^{-\frac{1}{2}}$

$\frac{d}{dx}[2x^{\color{DarkRed} \frac{1}{2}}]=\frac{1}{2}*\frac{1}{x^{\frac{1}{2}}}$

$\frac{d}{dx}[2x^{\color{DarkRed} \frac{1}{2}}]=\frac{1}{2\sqrt{x}}$ , first derivative

Now, solving the second derivative

$\frac{d}{dx}[\frac{1}{2\sqrt{x}}]=\frac{d}{dx}[\frac{1}{2}x^{{\color{DarkRed} -\frac{1}{2}}}]$

$\frac{d}{dx}[\frac{1}{2}x^{\color{DarkRed} 2}]={\color{DarkRed} -\frac{1}{2}}*\frac{1}{2}x^{{\color{DarkRed} -\frac{1}{2}}-1}$

$\frac{d}{dx}[\frac{1}{2}x^{\color{DarkRed} 2}]=-\frac{1}{4}x^{-\frac{3}{2}}$

$\frac{d}{dx}[\frac{1}{2}x^{\color{DarkRed} 2}]=-\frac{1}{4}*\frac{1}{x^{\frac{3}{2}}}$

$\frac{d}{dx}[\frac{1}{2}x^{\color{DarkRed} 2}]=-\frac{1}{4x^{\frac{3}{2}}}$

$\frac{d}{dx}[\frac{1}{2}x^{\color{DarkRed} 2}]=-\frac{1}{4\sqrt{x^3}}$ , second derivative (answer)

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$5.)\ \frac{3}{4}x^{\frac{4}{3}}$

$\frac{3}{4}x^{\frac{4}{3}}\ @{\color{DarkRed} \ n=\frac{4}{3}}$

$\frac{d}{dx}[\frac{3}{4}x^{{\color{DarkRed} \frac{4}{3}}}]={\color{DarkRed} \frac{4}{3}}*\frac{3}{4}x^{{\color{DarkRed} \frac{4}{3}}-1}$

$\frac{d}{dx}[\frac{3}{4}x^{{\color{DarkRed} \frac{4}{3}}}]=x^{\frac{4-3}{3}}$

$\frac{d}{dx}[\frac{3}{4}x^{{\color{DarkRed} \frac{4}{3}}}]=x^{\frac{1}{3}}$ , first derivative

Now, solving the second derivative

$\frac{d}{dx}[x^{{\color{DarkRed} \frac{1}{3}}}]={\color{DarkRed} \frac{1}{3}}x^{\frac{1}{3}-1}$

$\frac{d}{dx}[x^{{\color{DarkRed} \frac{1}{3}}}]={\color{DarkRed} \frac{1}{3}}x^{\frac{1-3}{3}$

$\frac{d}{dx}[x^{{\color{DarkRed} \frac{1}{3}}}]={\color{DarkRed} \frac{1}{3}}x^{-\frac{2}{3}$

$\frac{d}{dx}[x^{{\color{DarkRed} \frac{1}{3}}}]={\color{DarkRed} \frac{1}{3}}*\frac{1}{x^{\frac{2}{3}}}$

$\frac{d}{dx}[x^{{\color{DarkRed} \frac{1}{3}}}]={\color{DarkRed} \frac{1}{3}}*\frac{1}{\sqrt[3]{x^{2}}}$

$\frac{d}{dx}[x^{{\color{DarkRed} \frac{1}{3}}}]=\frac{1}{3\sqrt[3]{x^{2}}}$ , second derivative (answer)

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$6.)\ \frac{2}{x^{3}}$

$\frac{2}{x^{3}}\ @\ n=3$

$\frac{d}{dx}[\frac{2}{x^{3}}]=2x^{-3}$

$\frac{d}{dx}[\frac{2}{x^{3}}]=-3*2x^{-3-1}$

$\frac{d}{dx}[\frac{2}{x^{3}}]=-6x^{-4}$

$\frac{d}{dx}[\frac{2}{x^{3}}]=\frac{-6}{x^{4}}$ , first derivative

Now, solving the second derivative

$\frac{d}{dx}[-\frac{6}{x^{4}}]=-6x^{-4}$

$\frac{d}{dx}[-\frac{6}{x^{4}}]=-4*-6x^{-4-1}$

$\frac{d}{dx}[-\frac{6}{x^{4}}]=24x^{-5}$

$\frac{d}{dx}[-\frac{6}{x^{4}}]=\frac{24}{x^{5}}$ , second derivative (answer)

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$7.)\ \frac{3}{\sqrt{x}}$

$\frac{3}{\sqrt{x}}=3x^{-\frac{1}{2}}\ @\ {\color{DarkRed} n=-\frac{1}{2}}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=\frac{3}{x^{\frac{1}{2}}}=3x^{-\frac{1}{2}}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=-\frac{1}{2}*3x^{-\frac{1}{2}-1}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=-\frac{3}{2}x^{-\frac{3}{2}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=-\frac{3}{2}*\frac{1}{x^{\frac{3}{2}}}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=-\frac{3}{2}*\frac{1}{\sqrt{x^{3}}}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=-\frac{3}{2}{\color{DarkRed} \frac{1}{\sqrt{x^{2}}}}\frac{1}{\sqrt{x}}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=-\frac{3}{2}{\color{DarkRed} \frac{1}{x}}\frac{1}{\sqrt{x}}$

$\frac{d}{dx}[\frac{3}{\sqrt{x}}]=-\frac{3}{2x\sqrt{x}}$ , first derivative

Now, solving the second derivative

${\color{DarkBlue} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]}=-\frac{3}{2}\frac{1}{xx^{\frac{1}{2}}}=-\frac{3}{2}*\frac{1}{x^{\frac{3}{2}}}={\color{DarkBlue} -\frac{3x^{-\frac{3}{2}}}{2}}$

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]={\color{DarkBlue} -\frac{3x^{-\frac{3}{2}}}{2}}$ , rearranged as preparation

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]=-\frac{3}{2}({\color{DarkBlue} -\frac{3x^{-\frac{3}{2}-1}}{2}})$

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]=\frac{9x^{-\frac{5}{2}}}{4}$

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]=\frac{9}{4x^{\frac{5}{2}}}$

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]=\frac{9}{4\sqrt{x^{5}}}$

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]=\frac{9}{4\sqrt{x^{2}}\sqrt{x^{2}}\sqrt{x}}$

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]=\frac{9}{4xx*\sqrt{x}}$

${\color{Black} \frac{d}{dx}[-\frac{3}{2x\sqrt{x}}]=\frac{9}{4x^{2}\sqrt{x}}$ , second derivative (answer)

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$8.)\ 2a\sqrt{x}$

$2a\sqrt{x}=2ax^{\frac{1}{2}},\ @\ {\color{DarkRed} n=\frac{1}{2}}$

$\frac{d}{dx}[2a\sqrt{x}]=2ax^{\frac{1}{2}}$

$\frac{d}{dx}[2a\sqrt{x}]=\frac{1}{2}*2ax^{\frac{1}{2}-1}$

$\frac{d}{dx}[2a\sqrt{x}]=ax^{-\frac{1}{2}}$

$\frac{d}{dx}[2a\sqrt{x}]=\frac{a}{x^\frac{1}{2}}$

$\frac{d}{dx}[2a\sqrt{x}]=\frac{a}{\sqrt{x}}$ , first derivative

Now, solving the second derivative

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=\frac{a}{x^{\frac{1}{2}}}=ax^{-\frac{1}{2}}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{1}{2}*ax^{-\frac{1}{2}-1}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{1}{2}*ax^{-\frac{3}{2}}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{1}{2}*\frac{a}{x^{\frac{3}{2}}}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{1}{2}*\frac{a}{\sqrt[]{x^{3}}}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{1}{2}\frac{a}{\sqrt{x^{2}}\sqrt{x}}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{1}{2}\frac{a}{{\color{DarkRed} \sqrt{x^{2}}}\sqrt{x}}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{1}{2}\frac{a}{x\sqrt{x}}$

$\frac{d}{dx}[\frac{a}{\sqrt{x}}]=-\frac{a}{2x\sqrt{x}}$ , second derivative (answer)

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$9.)\ -2x^{\frac{3}{2}}$

$\ -2x^{\frac{3}{2}}\ @\ {\color{DarkRed} n=\frac{3}{2}}$

$\frac{d}{dx}[-2x^{\frac{3}{2}}]={\color{DarkRed} \frac{3}{2}}*-2x^{{\color{DarkRed} \frac{3}{2}}-1}$

$\frac{d}{dx}[-2x^{\frac{3}{2}}]=-3x^{\frac{1}{2}}$

$\frac{d}{dx}[-2x^{\frac{3}{2}}]=-3\sqrt{x}$ , first derivative

Now, solving the second derivative

$\frac{d}{dx}[-3\sqrt{x}]=-3x^{\frac{1}{2}}$

$\frac{d}{dx}[-3\sqrt{x}]=\frac{1}{2}*-3x^{\frac{1}{2}-1}$

$\frac{d}{dx}[-3\sqrt{x}]=-\frac{3}{2}x^{-\frac{1}{2}}$

$\frac{d}{dx}[-3\sqrt{x}]=-\frac{3}{2x^{\frac{1}{2}}}$

$\frac{d}{dx}[-3\sqrt{x}]=-\frac{3}{2\sqrt{x}}$ , second derivative (answer)

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$10.)\ \frac{5}{\sqrt{2x}}$

$\frac{5}{\sqrt{2x}}=\frac{5}{(2x)^{\frac{1}{2}}}$

$\frac{5}{\sqrt{2x}}=\frac{5}{(2x)^{\frac{1}{2}}}=5(2x)^{-\frac{1}{2}}\ @\ {\color{DarkRed} n=-\frac{1}{2}}$

$\frac{d}{dx}[\frac{5}{\sqrt{2x}}]=5(2x)^{-\frac{1}{2}}$

$\frac{d}{dx}[\frac{5}{\sqrt{2x}}]=-\frac{1}{2}*5(2x)^{-\frac{1}{2}-1}$

$\frac{d}{dx}[\frac{5}{\sqrt{2x}}]=-\frac{5(2x)^{-\frac{3}{2}}}{2}$

$\frac{d}{dx}[\frac{5}{\sqrt{2x}}]=-\frac{5}{2(2x)^{\frac{1}{2}}}$

$\frac{d}{dx}[\frac{5}{\sqrt{2x}}]=-\frac{5}{2\sqrt{2x}}$ , second derivative (answer)

Well, we hoped that you gain more ideas on how to do the second derivation on calculus. This is just the basic. Follow us so that you'll get connected as soon as we publish a new discussion like this. We appreciate your time and desire to know this indispensable mathematics calculus! Our TEAM is planning to answer textbooks-based problems so you guys have a better guide towards your journey!

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