Find the second order derivatives of the functions
(i) ${{x}^{2}}+3x+2$
(ii) ${{x}^{20}}$
Answer
631.2k+ views
Hint: We first assume the given functions as $y=f\left( x \right)$. We then define the second order derivatives as $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]=\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left[ f\left( x \right) \right]$. We use different differential theorems of power or indices value terms. We have the differentiation of constant terms as 0. Using the theorem and differentiating the given function twice we get the required solution.
Complete step by step answer:
We have to find the second order derivatives of the functions. We consider the functions as $y=f\left( x \right)$. We apply different differential theorems like $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.
We finally find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]=\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left[ f\left( x \right) \right]$.
For the first function $y=f\left( x \right)={{x}^{2}}+3x+2$. Applying the theorem, we get
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}+3x+2 \right) \\
& =\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( 3x \right)+\dfrac{d}{dx}\left( 2 \right) \\
& =2x+3 \\
\end{align}$
Now we have \[{{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=2x+3\]. We need to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ 2x+3 \right]=\dfrac{d}{dx}\left( 2x \right)+\dfrac{d}{dx}\left( 3 \right)=2$.
So, the second order derivative of ${{x}^{2}}+3x+2$ is 2.
For the second function $y=f\left( x \right)={{x}^{20}}$. Applying the theorem, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{20}} \right)=20{{x}^{19}}$
Now we have \[{{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=20{{x}^{19}}\]. We need to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ 20{{x}^{19}} \right]=\left( 20\times 19 \right){{x}^{19-1}}=380{{x}^{18}}$.
So, the second order derivative of ${{x}^{20}}$ is $380{{x}^{18}}$.
Note: We have the differentiation of power terms. The indices values can be anything belonging to real numbers. We can’t get to the second order derivatives directly. We have to differentiate twice every time.
Complete step by step answer:
We have to find the second order derivatives of the functions. We consider the functions as $y=f\left( x \right)$. We apply different differential theorems like $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.
We finally find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]=\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left[ f\left( x \right) \right]$.
For the first function $y=f\left( x \right)={{x}^{2}}+3x+2$. Applying the theorem, we get
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}+3x+2 \right) \\
& =\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( 3x \right)+\dfrac{d}{dx}\left( 2 \right) \\
& =2x+3 \\
\end{align}$
Now we have \[{{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=2x+3\]. We need to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ 2x+3 \right]=\dfrac{d}{dx}\left( 2x \right)+\dfrac{d}{dx}\left( 3 \right)=2$.
So, the second order derivative of ${{x}^{2}}+3x+2$ is 2.
For the second function $y=f\left( x \right)={{x}^{20}}$. Applying the theorem, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{20}} \right)=20{{x}^{19}}$
Now we have \[{{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=20{{x}^{19}}\]. We need to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ 20{{x}^{19}} \right]=\left( 20\times 19 \right){{x}^{19-1}}=380{{x}^{18}}$.
So, the second order derivative of ${{x}^{20}}$ is $380{{x}^{18}}$.
Note: We have the differentiation of power terms. The indices values can be anything belonging to real numbers. We can’t get to the second order derivatives directly. We have to differentiate twice every time.
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