Answer

Verified

435k+ views

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.

__Complete step-by-step answer:__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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is pollution? How many types of pollution? Define it