Question

# Find $\dfrac{dy}{dx}$, if ${{x}^{3}}+{{y}^{2}}+xy=10$.

Hint: To solve this question, differentiate the whole equation that is given in the question with respect to x. Use the product rule of derivative to differentiate the terms that are having two multiplied variables. To differentiate the functions of y, use chain rule. By this, we can find the $\dfrac{dy}{dx}$.

Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In differentiation, if we are required to differentiate a function f(x) which is equal to the product of other two functions g(x) and h(x) i.e. f(x) = g(x) $\times$ h(x), then the derivative of f(x) if given by the product rule of derivative. Hence, the derivative of f(x) by the product rule is given by the formula,
$f'\left( x \right)=g'\left( x \right).h\left( x \right)+g\left( x \right).h'\left( x \right).......................\left( 1 \right)$
In differentiation, if we are required to differentiate a function of y with respect to x, we use chain rule. The derivative of function of y using chain rule is given by,
$\dfrac{df\left( y \right)}{dx}=\left( \dfrac{df\left( y \right)}{dy} \right)\left( \dfrac{dy}{dx} \right)...........\left( 2 \right)$
Also, in differentiation, the derivative of ${{x}^{n}}$ is given by the formula,
$\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}................\left( 3 \right)$
In the question, we are given ${{x}^{3}}+{{y}^{2}}+xy=10$ and we are required to find $\dfrac{dy}{dx}$.
Let us differentiate ${{x}^{3}}+{{y}^{2}}+xy=10$ with respect to x.
\begin{align} & \dfrac{d}{dx}\left( {{x}^{3}}+{{y}^{2}}+xy \right)=\dfrac{d\left( 10 \right)}{dx} \\ & \Rightarrow \dfrac{d}{dx}{{x}^{3}}+\dfrac{d}{dx}{{y}^{2}}+\dfrac{d}{dx}xy=\dfrac{d\left( 10 \right)}{dx} \\ \end{align}
Using formula $\left( 1 \right)$ to differentiate the term xy, we get,
$\dfrac{d}{dx}{{x}^{3}}+\dfrac{d}{dx}{{y}^{2}}+\dfrac{dx}{dx}.y+x\dfrac{dy}{dx}=\dfrac{d\left( 10 \right)}{dx}$
Using formula $\left( 2 \right)$ to differentiate ${{y}^{2}}$, we get,
$\dfrac{d}{dx}{{x}^{3}}+\dfrac{d{{y}^{2}}}{dy}.\dfrac{dy}{dx}+\dfrac{dx}{dx}.y+x\dfrac{dy}{dx}=\dfrac{d\left( 10 \right)}{dx}$
Using formula $\left( 3 \right)$ to differentiate ${{x}^{3}}$ and ${{y}^{2}}$ , we get,
$3{{x}^{2}}+2y\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=\dfrac{d\left( 10 \right)}{dx}$
Since 10 is a constant, $\dfrac{d\left( 10 \right)}{dx}$ = 0. So, we get,
\begin{align} & 3{{x}^{2}}+2y\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0 \\ & \Rightarrow 3{{x}^{2}}+y+\dfrac{dy}{dx}\left( 2y+x \right)=0 \\ & \Rightarrow \dfrac{dy}{dx}\left( 2y+x \right)=-\left( 3{{x}^{2}}+y \right) \\ & \Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( 3{{x}^{2}}+y \right)}{\left( 2y+x \right)} \\ \end{align}
Hence, the answer is $-\dfrac{\left( 3{{x}^{2}}+y \right)}{\left( 2y+x \right)}$.

Note: There is a possibility that one may commit a mistake while differentiating xy with respect to x. There is a possibility that one may forget to apply chain rule to the term xy in ${{x}^{3}}+{{y}^{2}}+xy=10$. This mistake will lead us to an incorrect answer.