Question

How do you differentiate $-5$?

Answer 1

From the question given we have to find the differentiation of $-5$. Differentiation means it is a method of finding the derivative of a function. Differentiation is a process, in maths, where we find the instantaneous rate of change in function based on its variables. The derivative of a function describes the functions instantaneous rate of change at a certain moment; the most common example is the rate change of displacement with respect to time is called velocity. In differentiation the derivative of constant function is zero. In the above question the given function is constant function i.e.,$-5$. So, the differentiation of the given constant function is zero.

Complete step-by-step solution:
From the question given we have to find the differentiation of $-5$. The given function is a constant function.
As we know that the differentiation of any constant function is zero.
So, the differentiation of the above function is zero,
$\Rightarrow \dfrac{d}{dx}\left( -5 \right)=0$
Therefore, the differentiation of the above function i.e., $-5$ is zero

Note: Student should know the formulas of differentiation. Student should know the basic formulas of differentiation like;

& \Rightarrow \dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}} \\\ & \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}=-\cos e{{c}^{2}}x \\\ & \Rightarrow \dfrac{d\left( \sec x \right)}{dx}=\sec x\times \tan x \\\ & \Rightarrow \dfrac{d\left( \cos ecx \right)}{dx}=-\cos ecx \times \cot x \\\ & \Rightarrow \dfrac{d\left( cons\tan t \right)}{dx}=0 \\\ \end{aligned}$$ Students should also know the UV rule. $$\Rightarrow D\left( UV \right)=D\left( U \right)\times V+D\left( V \right)\times U$$ student should also know about the U by V rule, if any function is in the form U by V then the differentiating that function will be done by U by V rule, $$\Rightarrow D\left( \dfrac{U}{V} \right)=\dfrac{V\times D\left( U \right)-U\times D\left( V \right)}{{{V}^{2}}}$$ . Students should not make any calculation mistakes.