Question

If $2{x^2} - 3xy + {y^2} + x + 2y - 8 = 0$, then ${{dy}}{{dx}}$ is equal to
A.${{\left( {3y - 4x - 1} \right)}}{{\left( {2y - 3x + 2} \right)}}$
B.${{\left( {3y + 4x + 1} \right)}}{{\left( {2y + 3x + 2} \right)}}$
C.${{\left( {3y - 4x + 1} \right)}}{{\left( {2y - 3x - 2} \right)}}$
D.${{\left( {3y - 4x + 1} \right)}}{{\left( {2y + 3x + 2} \right)}}$

Answer 1

First, we shall analyze the given information so that we can able to solve the problem. Generally in Mathematics, the derivative refers to the rate of change of a function with respect to a variable. Here in this question, we are asked to calculate the first derivative of the given equation. First, we need to split the given equation for our convenience. Then, we need to differentiate the resultant equation. Here, we are applying the power rule of differentiation to find the required answer.
Formula to be used:
The formula for the power rule of differentiation is as follows.
${d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step answer:
It is given that
$2{x^2} - 3xy + {y^2} + x + 2y - 8 = 0$
To find:${{dy}}{{dx}}$
Let us consider $2{x^2}$
Then ${d}{{dx}}\left( {2{x^2}} \right) = {{2d{x^2}}}{{dx}}$
$= 2 \times 2{x^{2 - 1}}$ (Here we applied the power rule of differentiation)
$= 4x$
Let us consider $- 3xy$
Then ${d}{{dx}}\left( { - 3xy} \right) = - 3{d}{{dx}}\left( {xy} \right)$
$= - 3\left( {x{{dy}}{{dx}} + y} \right)$ (In this step, we used the product rule of differentiation)
$= - 3x{{dy}}{{dx}} - 3y$
Let us consider ${y^2}$
Then ${d}{{dx}}\left( {{y^2}} \right) = 2y{{dy}}{{dx}}$
Now, let us consider $x$
Then ${d}{{dx}}\left( x \right) = 1 \times {x^{\left( {1 - 1} \right)}}$ (In this step, we applied the power rule of differentiation)
$= 1 \times x^\circ$
$= 1$
Now, let us consider $2y$
Then ${d}{{dx}}\left( {2y} \right) = 2{{dy}}{{dx}}$
We all know that the derivative of constant is zero.
Hence ${d}{{dx}}\left( { - 8} \right) = 0$
The given equation is $2{x^2} - 3xy + {y^2} + x + 2y - 8 = 0$
Now, differentiate the given equation with respect to $n$
That is ${d}{{dx}}\left( {2{x^2} - 3xy + {y^2} + x + 2y - 8} \right) = 0$
We shall separate the terms.
Thus ${d}{{dx}}\left( {2{x^2}} \right) + {d}{{dx}}\left( { - 3xy} \right) + {d}{{dx}}\left( {{y^2}} \right) + {d}{{dx}}\left( x \right) + {d}{{dx}}\left( {2y} \right) + {d}{{dx}}\left( { - 8} \right) = 0$
Now we need to substitute the obtained result in the above equation.
Hence we get $4x - 3x{{dy}}{{dx}} - 3y + 2y{{dy}}{{dx}} + 1 + 2{{dy}}{{dx}} + 0 = 0$
$(4x - 3y + 1) + {{dy}}{{dx}}(2x - 3x + 2) = 0$
${{dy}}{{dx}}(2y - 3x + 2) = - \left( {4x - 3y + 1} \right)$
$= 3y - 4x - 1$
${{dy}}{{dx}} = {{3y - 4x - 1}}{{2y - 3x + 2}}$

Thus option $\left( A \right)$ is correct answer.**

Note:
We often use the power rule to calculate the derivative of a variable raised to a power and the power rule is the most commonly used derivative rule. When we are asked to find the derivation of the given equation, we need to change the given equation for our convenience. Then we need to analyze that where we need to apply the derivative formulae and where we need to apply the rule of differentiation while differentiating the given equation.