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Question: What is the derivative of \(y=2{{x}^{2}}-5\)?...

What is the derivative of y=2x2−5y=2{{x}^{2}}-5?

Explanation

Solution

From the question given we have been asked to find the derivative of the y=2x2−5y=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 xn{{x}^{n}} is n×xn−1n\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
⇒y=2x2−5\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,
⇒dydx=2d(x2)dx−d(5)dx\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
⇒dydx=2d(x2)dx−0\Rightarrow \dfrac{dy}{dx}=2\dfrac{d\left( {{x}^{2}} \right)}{dx}-0
Now we know that the differentiation of xn{{x}^{n}} is n×xn−1n\times {{x}^{n-1}}
From this we will get,
⇒dydx=2×2×x2−1\Rightarrow \dfrac{dy}{dx}=2\times 2\times {{x}^{2-1}}
By further simplification we will get,
⇒dydx=4x\Rightarrow \dfrac{dy}{dx}=4x
Therefore, the derivative of y=2x2−5y=2{{x}^{2}}-5 is 4x4x.

Note: Students should know the formulas clearly if in the above question if students write derivative of 2x22{{x}^{2}}is 2×x2−122\times \dfrac{{{x}^{2-1}}}{2} instead of 2×2x2−12\times 2{{x}^{2-1}} the whole solution will be wrong. So, Students should know the 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}