Question

What is the derivative of $y=2{{x}^{2}}-5$?

Answer 1

Hint: From the question given we have been asked to find the derivative of the $y=2{{x}^{2}}-5$. As we know that the basic formulas of differentiation like, derivative of any constant is equal to zero and derivative of ${{x}^{n}}$ is $n\times {{x}^{n-1}}$. By these formulas we will get the required answer.

Complete step by step solution:
From the question given that we have to find the derivative of
$\Rightarrow y=2{{x}^{2}}-5$
Now we have to do differentiation on both sides with respect to x,
By doing differentiation on both sides with respect to x we will get,
$\Rightarrow \dfrac{dy}{dx}=2\dfrac{d\left( {{x}^{2}} \right)}{dx}-\dfrac{d\left( 5 \right)}{dx}$
As we know that the basic formulas of differentiation like, derivative of any constant is equal to zero.
From this the differentiation of constant 5 is 0, that is
$\Rightarrow \dfrac{dy}{dx}=2\dfrac{d\left( {{x}^{2}} \right)}{dx}-0$
Now we know that the differentiation of ${{x}^{n}}$ is $n\times {{x}^{n-1}}$
From this we will get,
$\Rightarrow \dfrac{dy}{dx}=2\times 2\times {{x}^{2-1}}$
By further simplification we will get,
$\Rightarrow \dfrac{dy}{dx}=4x$
Therefore, the derivative of $y=2{{x}^{2}}-5$ is $4x$.

Note: Students should know the formulas clearly if in the above question if students write derivative of $2{{x}^{2}}$is $2\times \dfrac{{{x}^{2-1}}}{2}$ instead of $2\times 2{{x}^{2-1}}$ the whole solution will be wrong. So, Students should know the basic formulas of differentiation like,
$\begin{align}
  & \Rightarrow \dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}} \\
 & \Rightarrow \dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x} \\
 & \Rightarrow \dfrac{d\left( \sin x \right)}{dx}=\cos x \\
 & \Rightarrow \dfrac{d\left( \cos x \right)}{dx}=-\sin x \\
 & \Rightarrow \dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x \\
 & \Rightarrow \dfrac{d\left( \cot x \right)}{dx}={{\operatorname{cosec}}^{2}}x \\
 & \Rightarrow \dfrac{d\left( \sec x \right)}{dx}=\sec x\times \tan x \\
 & \Rightarrow \dfrac{d\left( \operatorname{cosec}x \right)}{dx}=-\operatorname{cosec}x\times \cot x \\
 & \Rightarrow \dfrac{d\left( cons\tan t \right)}{dx}=0 \\
\end{align}$