Question

Find the derivative of function \[12{{x}^{2}}-31\] with respect to x?

Answer 1

Hint: We start solving the problem by applying derivative with respect to x for the given function. We then use the formula $\dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)-\dfrac{d}{dx}\left( g\left( x \right) \right)$ to proceed through the problem. We then use the facts that derivative of a constant is zero and $\dfrac{d}{dx}\left( af\left( x \right) \right)=a\dfrac{d}{dx}\left( f\left( x \right) \right)$ to proceed further through the problem. We then use the formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\left( {{x}^{n-1}} \right)$ and make necessary calculations in order to get the required result.

Complete step-by-step answer:
According to the problem, we need to find the derivative of the function \[12{{x}^{2}}-31\] with respect to x.
Let us apply the derivative with respect to x for the function \[12{{x}^{2}}-31\].
So, we have $\dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)$ ---(1).
We know that $\dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)-\dfrac{d}{dx}\left( g\left( x \right) \right)$, we use this result in equation (1).
$\Rightarrow \dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)=\dfrac{d}{dx}\left( 12{{x}^{2}} \right)-\dfrac{d}{dx}\left( 31 \right)$ ---(2).
We know that the derivative of any constant ‘a’ is zero i.e., $\dfrac{d}{dx}\left( a \right)=0$ and $\dfrac{d}{dx}\left( af\left( x \right) \right)=a\dfrac{d}{dx}\left( f\left( x \right) \right)$. We use these results in equation (2).
$\Rightarrow \dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)=12\dfrac{d}{dx}\left( {{x}^{2}} \right)-0$.
$\Rightarrow \dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)=12\dfrac{d}{dx}\left( {{x}^{2}} \right)$ ---(3).
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\left( {{x}^{n-1}} \right)$. We use this result in equation (3).
$\Rightarrow \dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)=12\left( 2{{x}^{2-1}} \right)$.
$\Rightarrow \dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)=12\left( 2{{x}^{1}} \right)$.
$\Rightarrow \dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)=12\left( 2x \right)$.
$\Rightarrow \dfrac{d}{dx}\left( 12{{x}^{2}}-31 \right)=24x$.
So, we have the derivative of the function $12{{x}^{2}}-31$ as $24x$.
∴ The derivative of the function $12{{x}^{2}}-31$ is $24x$.

Note: We should not confuse the formulas of derivatives. We should not make calculation mistakes while solving this problem. Alternatively, we can solve this problem as follows:
We know that the derivative of the function $f\left( x \right)$ is defined as $\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$.
Let us assume $f\left( x \right)=12{{x}^{2}}-31$.
So, we have \[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)\].
$\Rightarrow {{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{12{{\left( x+h \right)}^{2}}-31-\left( 12{{x}^{2}}-31 \right)}{h} \right)$.
$\Rightarrow {{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{12\left( {{x}^{2}}+{{h}^{2}}+2hx \right)-31-12{{x}^{2}}+31}{h} \right)$.
$\Rightarrow {{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{12{{x}^{2}}+12{{h}^{2}}+24hx-12{{x}^{2}}}{h} \right)$.
$\Rightarrow {{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{12{{h}^{2}}+24hx}{h} \right)$.
$\Rightarrow {{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 12h+24x \right)$.
$\Rightarrow {{f}^{'}}\left( x \right)=12\left( 0 \right)+24x$.
$\Rightarrow {{f}^{'}}\left( x \right)=0+24x$.
$\Rightarrow {{f}^{'}}\left( x \right)=24x$.
So, the derivative of the function $12{{x}^{2}}-31$ is $24x$.