Question

How do you calculate $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ of $y=-4{{x}^{2}}+7x$?

Answer 1

Consider the given function and differentiate both the sides with respect to x to find the first derivative, i.e., $\dfrac{dy}{dx}$. Apply the formula: - $\dfrac{d\left[ {{x}^{a}} \right]}{dx}=a.{{x}^{a-1}}$ to simplify the derivative. Now, again differentiate the function, obtained from first derivative, with respect to x on both the sides to find the second derivative, i.e., $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. Use the property that “derivative of a constant term is 0” and get the required expression.

Complete step by step answer:
Here, we have been provided with the function, $y=-4{{x}^{2}}+7x$, and we are asked to find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$, i.e., the second derivative of the function. But first we need to find the first derivative of y, i.e., $\dfrac{dy}{dx}$.
Now, we can see that the function y contains the variable x, so differentiating both the sides with respect to x, we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( -4{{x}^{2}}+7x \right)}{dx}$
Breaking the terms on the Right-hand side, we have,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( -4{{x}^{2}} \right)}{dx}+\dfrac{d\left( 7x \right)}{dx}$
Now, in the first term (-4) is a constant multiplied to the variable ${{x}^{2}}$ and in the second term 7 is a constant multiplied to the variable x, so taking these constants out of the derivative expression, we get,
$\Rightarrow \dfrac{dy}{dx}=\left( -4 \right)\dfrac{d{{x}^{2}}}{dx}+7\dfrac{dx}{dx}$
Applying the formula: - $\dfrac{d\left[ {{x}^{a}} \right]}{dx}=a.{{x}^{a-1}}$, we get,
$\Rightarrow \dfrac{dy}{dx}=-4\times 2{{x}^{1}}\dfrac{dx}{dx}+7$
$\Rightarrow \dfrac{dy}{dx}=-8x+7$
We have found the first derivative, now to find the second derivative let us differentiate the above relation again with respect to x. So, we have,

& \Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d\left( -8x+7 \right)}{dx} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( -8x+7 \right)}{dx} \\\ \end{aligned}$$ Breaking the terms on the right-hand side, we have, $$\begin{aligned} & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( -8x \right)}{dx}+\dfrac{d\left( 7 \right)}{dx} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-8\left( \dfrac{dx}{dx} \right)+\dfrac{d\left( 7 \right)}{dx} \\\ \end{aligned}$$ Clearly, we can see that in the R.H.S 7 is a constant and we know that the derivative of a constant term is 0. So, we have, $$\begin{aligned} & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-8+0 \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-8 \\\ \end{aligned}$$ Hence, the above relation denotes the second derivative of the function y. **Note:** One must remember the basic formulas of differentiation of common functions because they are used everywhere. Always remember that the derivative of a constant term is always 0. We cannot directly find the second derivative of the given function, first we need to find the first derivative. In higher classes we do have the formula for $${{n}^{th}}$$ differentiation of a function but here we are not going to use that. So, proceed step by step as above.