Question

How do you find the derivative of ${{x}^{2}}+xy+{{y}^{2}}=7?$

Answer 1

Hint: We will first recall the concept of differentiation and its various method to solve and then we will find the differentiation of ${{x}^{2}}+xy+{{y}^{2}}=7$. To find the differentiation of a given explicit function. we will differentiate the given function both sides with respect to x using the product and sum rule of derivative of two different function and then we will use the standard differentiation formulas such as $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ and we also know that derivative of constant is zero so we will also use this property.

Complete step-by-step solution:
We will use the concept of methods of differentiation to solve the above question. We can see from the question that the given function is explicit as it has multiplication of both x and y together and they are not separated so to find the derivative of the above given function we will first simply differentiate the given equation with respect to x.
We will apply the product rule and sum rule of differentiation for two different functions.
Let $f\left( x \right)$ and $g\left( x \right)$ be two differentiable function, then differentiation of their product is given as:
$\dfrac{d\left( f\left( x \right).g\left( x \right) \right)}{dx}=f\left( x \right)\times \dfrac{d\left( g\left( x \right) \right)}{dx}+\dfrac{d\left( f\left( x \right) \right)}{dx}\times g\left( x \right)$
Also, the differentiation of their sum is given as:
$\dfrac{d\left( f\left( x \right)+g\left( x \right) \right)}{dx}=\dfrac{d\left( f\left( x \right) \right)}{dx}+\dfrac{d\left( g\left( x \right) \right)}{dx}$
So, after differentiating both sides the given equation ${{x}^{2}}+xy+{{y}^{2}}=7$, we will get:
$\Rightarrow \dfrac{d\left( {{x}^{2}}+xy+{{y}^{2}} \right)}{dx}=\dfrac{d\left( 7 \right)}{dx}$
Since, we know that derivative of constant is zero and after applying the sum rule of differentiation we will get:
\[\Rightarrow \dfrac{d\left( {{x}^{2}} \right)}{dx}+\dfrac{d\left( xy \right)}{dx}+\dfrac{d\left( {{y}^{2}} \right)}{dx}=0\]
Now, we know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, and we will also apply the product rule of so we will get:
\[\Rightarrow 2{{x}^{2-1}}+y\dfrac{d\left( x \right)}{dx}+x\dfrac{d\left( y \right)}{dx}+2y\dfrac{d\left( y \right)}{dx}=0\]
\[\Rightarrow 2x+y\times 1+x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0\]
Now, we will take $\dfrac{dy}{dx}$ common from the last two term, then we will get:
\[\Rightarrow 2x+y+\left( x+2y \right)\dfrac{dy}{dx}=0\]
\[\Rightarrow \left( x+2y \right)\dfrac{dy}{dx}=-\left( 2x+y \right)\]
Now, after dividing both side by $\left( x+2y \right)$ we will get:
\[\Rightarrow \dfrac{\left( x+2y \right)}{\left( x+2y \right)}\dfrac{dy}{dx}=-\dfrac{\left( 2x+y \right)}{\left( x+2y \right)}\]
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( 2x+y \right)}{\left( x+2y \right)}\]
Hence derivative of the ${{x}^{2}}+xy+{{y}^{2}}=7$ is \[\dfrac{dy}{dx}=-\dfrac{\left( 2x+y \right)}{\left( x+2y \right)}\]
This is our required solution.

Note: Students are required to memorize all the standard derivative formulas otherwise they will not be able to solve any question. Also, note that the product rule can be extended to more than two functions. Let us say that $f\left( x \right),g\left( x \right)\text{ and }h\left( x \right)$ are three differentiable function, then differentiation of their product is given as:
$\dfrac{d\left( f\left( x \right).g\left( x \right).h\left( x \right) \right)}{dx}=f\left( x \right).g\left( x \right).\dfrac{d\left( h\left( x \right) \right)}{dx}+f\left( x \right).\dfrac{d\left( g\left( x \right) \right)}{dx}.h\left( x \right)+\dfrac{d\left( f\left( x \right) \right)}{dx}.g\left( x \right).h\left( x \right)$ .
Similarly, we can write for more than 3 functions.