Question

If ${{x}^{3}}+8xy+{{y}^{3}}=64$ then find $\dfrac{dy}{dx}$ ?
1. $\dfrac{-(3{{x}^{2}}+8y)}{(8x+3{{y}^{2}})}$
2. $\dfrac{(3{{x}^{2}}+8y)}{(8x+3{{y}^{2}})}$
3. $\dfrac{(3x+8y)}{(8{{x}^{2}}+3y)}$
4. None of these

Answer 1

Hint : In this question we can see that $\dfrac{dy}{dx}$ means that we need to take the differentiation of the equation given to us with respect to x. And then we need to know the formulas and logics of differentiation to be able to solve this question.

Complete step-by-step answer :
Now here we can start the solution by knowing the expression whose derivative we need to find. Here in this case the derivative we need to find the of the function
${{x}^{3}}+8xy+{{y}^{3}}=64$
Now we need to find the derivative of the function, so to do that we need to use different formulas for different terms in this question. As we can say for the first term we have the cube of x so here to solve it we use the formula that $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ . Now likewise for the second term we can see that here there are two different variables since the constant is taken out during differentiation since its derivation is zero. Now since we have two variable we can use the product rule which goes like
$\dfrac{duv}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ . Now in the last term we can see that it is similar to the first term so here too you can solve it using the similar formula. In the RHS of the equation we can see that there is only a constant so taking a derivative of a constant will give us zero. All of these derivatives need to be with respect to x to get the correct answer. Therefore taking the derivative of the expression
$\dfrac{d({{x}^{3}})}{dx}+\dfrac{d(8xy)}{dx}+\dfrac{d{{y}^{3}}}{dx}=\dfrac{d(64)}{dx}$
Using the formulas we get
$3{{x}^{3-1}}+8x\dfrac{dy}{dx}+8y\dfrac{dx}{dx}+3{{y}^{3-1}}\dfrac{dy}{dx}=0$
Now we know that $\dfrac{dx}{dx}=1$ therefore
$3{{x}^{2}}+8x\dfrac{dy}{dx}+8y+3{{y}^{2}}\dfrac{dy}{dx}=0$
Now taking the terms which do not have $\dfrac{dy}{dx}$ on the right hand side
$+8x\dfrac{dy}{dx}+3{{y}^{2}}\dfrac{dy}{dx}=-(3{{x}^{2}}+8y)$
Taking $\dfrac{dy}{dx}$ common on LHS of equation
$\dfrac{dy}{dx}(8x+3{{y}^{2}})=-(3{{x}^{2}}+8y)$
Cross multiplying we get
$\dfrac{dy}{dx}=\dfrac{-(3{{x}^{2}}+8y)}{(8x+3{{y}^{2}})}$
Hence we can say that the derivative of the equation is $\dfrac{-(3{{x}^{2}}+8y)}{(8x+3{{y}^{2}})}$ and the answer for this question is option one.
So, the correct answer is “Option 1”.

Note : Now in questions where we are asked for derivatives a common mistake observed is that some may instead of taking y as another derivative end up taking y as a constant and that might get us the wrong answer. Also whichever variable is at the bottom of the derivative let it be y or x than that variable is the variable with respect to whom we will find the derivative