Question

How do we find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ if $({x^3})y - x - 2y - 6 = 0$ ?

Answer 1

To solve this question, we will use the implicit type of differentiation in which we will find the derivative until we achieve the $\dfrac{dy}{dx}$ situation. That’s how we can solve the given equation in the terms of $x$ and $y$ .

Complete step by step answer:
For this type question, we use implicit differentiation (basically regular differentiation, but with $y$ as well as $x$). It works the same way as regular differentiation, except every time we have to take the derivative of $y$ , we have to put ${y'}$ after it. For the first term, we have to use the product rule, which is:
$({f'}(X) \times g(x)) + (f(x) \times {g'}(x))$
Here $f(x) = {x^3}$ and $g(x) = y$ . So we take the derivative of ${x^3}$ and multiply by $y$ , then switch, so we get the first term (surrounded by parenthesis in the answer).
The rest is quite simple, the derivative of $x$ is 1, the derivative of $- 2y$ is $- 2{y'}$ and 6 and 0 are both 0. We put all together, and we get:
$(3{x^2}y + {x^3}{y'}) - 1 - 2{y'} = 0$

Note: Since the functions cannot be expressed in terms of one specific variable, we have to follow a different method to find the derivative of the implicit function: While computing the derivative of the Implicit function, our aim is to solve for $\dfrac{dy}{dx}$ or any higher-order derivatives depending on the function.