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Question: What is the derivative of \(4{{\pi }^{2}}\) ?...

What is the derivative of 4π24{{\pi }^{2}} ?

Explanation

Solution

From the question given that we have to find the derivative of 4π24{{\pi }^{2}}. As we know that the definition of a derivative it is the instantaneous rate of change. As we know that the derivative of a constant is zero because it does not vary. The question given function is a constant that is 4π24{{\pi }^{2}} because π\pi has a value 3.14 or 227\dfrac{22}{7}. So, the derivative of this is zero.

Complete step-by-step solution:
From the question given that we have to find the derivative of
⇒4π2\Rightarrow 4{{\pi }^{2}}
As we know that the definition of a derivative is the instantaneous rate of change or derivatives are defined as the varying rate of change of function with respect to an independent variable. The derivative is mainly used when a quantity varies, and the rate of change is not constant.
According to the definition of a derivative. The derivative of a constant is zero because it doesn’t vary. So, the rate of change of constant is zero.
In the given question the function 4π24{{\pi }^{2}} is a constant because π\pi has a value 3.14 or 227\dfrac{22}{7}. so, the 4π24{{\pi }^{2}}is a constant. As we know that the derivative of constant is zero.
⇒d(4π2)dx=0\Rightarrow \dfrac{d\left( 4{{\pi }^{2}} \right)}{dx}=0
Therefore, the derivative of 4π24{{\pi }^{2}} is zero.

Note: Students should know that if in place of π\pi there is a variable x then the derivative of 4x24{{x}^{2}} is
⇒d(4x2)dx=8x\Rightarrow \dfrac{d\left( 4{{x}^{2}} \right)}{dx}=8x. Students should know some basic formulas of differentiation like,
⇒d(xn)dx=n×xn−1 ⇒d(log⁡x)dx=1x ⇒d(sin⁡x)dx=cos⁡x ⇒d(cos⁡x)dx=−sin⁡x ⇒d(tan⁡x)dx=sec⁡2x ⇒d(cot⁡x)dx=cosec⁡2x ⇒d(sec⁡x)dx=sec⁡x×tan⁡x ⇒d(cosec⁡x)dx=−cosec⁡x×cot⁡x ⇒d(constan⁡t)dx=0 \begin{aligned} & \Rightarrow \dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}} \\\ & \Rightarrow \dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x} \\\ & \Rightarrow \dfrac{d\left( \sin x \right)}{dx}=\cos x \\\ & \Rightarrow \dfrac{d\left( \cos x \right)}{dx}=-\sin x \\\ & \Rightarrow \dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x \\\ & \Rightarrow \dfrac{d\left( \cot x \right)}{dx}={{\operatorname{cosec}}^{2}}x \\\ & \Rightarrow \dfrac{d\left( \sec x \right)}{dx}=\sec x\times \tan x \\\ & \Rightarrow \dfrac{d\left( \operatorname{cosec}x \right)}{dx}=-\operatorname{cosec}x\times \cot x \\\ & \Rightarrow \dfrac{d\left( cons\tan t \right)}{dx}=0 \\\ \end{aligned}