Question

What is the second derivative of ${{x}^{2}}+{{y}^{2}}=25$ evaluated at $\left( -3,-4 \right)$ ?

Answer 1

We are given an equation of second degree of which we are supposed to find the second derivative by applying simple formulas of derivation We first differentiate the entire equation with respect to $x$ . Upon doing that we evaluate the first derivative and again differentiate it with respect to $x$ which gives us the second derivative. Then, substituting the value of $x$ and $y$ we reach to the answer of the given problem.

Complete step-by-step solution:
The equation we are given is ${{x}^{2}}+{{y}^{2}}=25$
We now differentiate the above equation with respect to $x$ using the sum rule i.e., ${{\left( a+b \right)}^{\prime }}={a}'+{b}'$ as shown below
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 25 \right)$
$\Rightarrow 2x+\dfrac{dy}{dx}\left( 2y \right)=0$
Further simplifying we get
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}$
Now we again differentiate the above equation with respect to $x$ by using the quotient rule of differentiation as shown below
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( -\dfrac{x}{y} \right)$
According to the quotient rule of differentiation $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\cdot {u}'-u\cdot {v}'}{{{v}^{2}}}$
Hence, we get the differentiation as
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{y\cdot 1-x\dfrac{dy}{dx}}{{{y}^{2}}}$
Now we substitute the value of the first derivative i.e., $\dfrac{dy}{dx}=-\dfrac{x}{y}$ in the above equation as shown below
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{y\cdot 1-x\left( -\dfrac{x}{y} \right)}{{{y}^{2}}}$
Further omitting the brackets and simplifying the above equation we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{y+\dfrac{{{x}^{2}}}{y}}{{{y}^{2}}}$
Multiplying both the denominator and numerator to $y$we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{y\cdot y+y\cdot \dfrac{{{x}^{2}}}{y}}{y\cdot {{y}^{2}}}$
Further doing some simplifications in the above equation we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{y}^{2}}+{{x}^{2}}}{{{y}^{3}}}$
We already know that ${{x}^{2}}+{{y}^{2}}=25$
Hence, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{25}{{{y}^{3}}}$
As, the above expression for second derivative only consists of the variable $y$ , we only put the value of $y$ as $-4$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{25}{{{\left( -4 \right)}^{3}}}$
Simplifying we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{25}{64}$
Therefore, we conclude that the second derivative of the required given is $\dfrac{25}{64}$ .

Note: While doing the differentiation process we have to be careful about applying the formulas so that they are used appropriately. Also, we must not keep any flaws while substituting the value of the first derivative to avoid errors in the solution.