Question

How do you find $\dfrac{{{d^2}y}}{{d{x^2}}}$ for $2x + {y^2} = 5$?

Answer 1

Differentiate both sides of the given function with respect to the variable $x$. Use the formula: - $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ to simplify the L.H.S. Use the chain rule of differentiation to find the derivative of ${y^2}$. In the R.H.S. use the factor that the derivative of a constant is 0 to simplify. Keep the expression $\dfrac{{dy}}{{dx}}$ in the L.H.S. and send the other variables and expressions to the R.H.S. to get the answer. Next, again differentiate both sides of the given function with respect to the variable $x$. Use the chain rule and value of $\dfrac{{dy}}{{dx}}$ to get the desired result.

Formula used:
(i). Chain Rule:
Chain rule is applied when the given function is the function of function i.e.,
if y is a function of x, then $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ or $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dv}} \times \dfrac{{dv}}{{dx}}$.
(ii). The differentiation of the product of a constant and a function = the constant $\times$ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
(iii). Power rule: $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},n \ne - 1$
(iv). Product rule: $\dfrac{d}{{dx}}\left( {fg} \right) = f\dfrac{d}{{dx}}g + g\dfrac{d}{{dx}}f$

Complete step-by-step solution:
Here, we have been provided with the relation: - $2x + {y^2} = 5$ and we are asked to find the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$.
$\because 2x + {y^2} = 5$
Now, differentiating both the sides of the above relation with respect to the variable $x$, we get,
$\dfrac{d}{{dx}}\left( {2x + {y^2}} \right) = \dfrac{d}{{dx}}\left( 5 \right)$
Breaking the terms in the L.H.S., we get,
$\Rightarrow 2\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( 5 \right)$
Now, using the formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$, we get,
$\Rightarrow 2 + \dfrac{{d{y^2}}}{{dx}} = \dfrac{d}{{dx}}\left( 5 \right)$
We know that the derivative of a constant term is 0, so in the R.H.S. we must have 0,
$\Rightarrow 2 + \dfrac{{d{y^2}}}{{dx}} = 0$
Applying the chain rule of differentiation to find the derivative of $\dfrac{{d{y^2}}}{{dx}}$, we have,
$\Rightarrow 2 + \dfrac{{d{y^2}}}{{dy}} \times \dfrac{{dy}}{{dx}} = 0$
What we are doing is, first we are differentiating ${y^2}$ with respect to $y$ and then we are differentiating $y$ with respect to $x$ and their product is considered. So, we have,
$\Rightarrow 2 + 2y\dfrac{{dy}}{{dx}} = 0$
Dividing both the sides with 2, we get,
$\Rightarrow 1 + y\dfrac{{dy}}{{dx}} = 0$
$\Rightarrow y\dfrac{{dy}}{{dx}} = - 1$
Dividing both the sides with $y$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{y}$…(1)
Now, again differentiating both the sides of the above relation with respect to the variable $x$, we get,
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = - \dfrac{d}{{dx}}\left( {\dfrac{1}{y}} \right)$
Applying the chain rule of differentiation to find the derivative of $\dfrac{{d{y^{ - 1}}}}{{dx}}$, we have,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{d{y^{ - 1}}}}{{dy}} \times \dfrac{{dy}}{{dx}}$
What we are doing is, first we are differentiating ${y^{ - 1}}$ with respect to $y$ and substitute the value of $\dfrac{{dy}}{{dx}}$ from (1) and their product is considered. So, we have,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {y^{ - 2}} \times \left( { - \dfrac{1}{y}} \right)$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = -\dfrac{1}{{{y^3}}}$
Therefore, $\dfrac{{{d^2}y}}{{d{x^2}}} = -\dfrac{1}{{{y^3}}}$.

Note: One may note that we can further simplify the relation that we have obtained, i.e., $\dfrac{{{d^2}y}}{{d{x^2}}} = -\dfrac{1}{{{y^3}}}$, by substituting the value of $y$ in terms of $x$. It can be written as: - $y = \sqrt {5 - 2x}$. In this way we will get the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ in terms of variable $x$. However, it is not of much importance here. You must remember the different rules of differentiation like: - the product rule, chain rule, $\dfrac{u}{v}$ rule etc. Because they are common rules and are used everywhere in calculus.