Question

How do you use implicit differentiation ${x^2} + 2xy - {y^2} + x = 2$ and to find an
equation of the tangent line to the curve , at the point $\left( {1,2} \right)$?

Answer 1

Hint: Start by considering $f(x)$ as the function of $x$. Next step is substitution. Substitute the values in place of the terms to make the equation easier to solve. Then we apply the chain rule to solve the derivative. We will solve the derivatives of each term separately. To find the slope of the tangent line of the curve passing $(1,2)$, put $x = 1\,and\,y = 2$ and for the equation use the slope point form $y - y_1 = m(x - x_1)$

Formula used:
Slope-Point form =$y - y_1 = m(x - x_1)$

Complete step by step solution:
We are given ${x^2} + 2xy - {y^2} + x = 2$ and we have to find its implicit differentiation with respect to $x$.
${x^2} + 2xy - {y^2} + x = 2$
Differentiating both sides of the equation with respect to $x$.
$\dfrac{d}{{dx}}\left( {{x^2} + 2xy - {y^2} + x} \right) = \dfrac{d}{{dx}}\left( 2 \right)$
Remember one thing the $y$in the above equation is the name of some other function of $x$,which we don’t know. $2xy$ is actually $2x.\left( {some\,function\,of\,x} \right)$
So, to differentiate $2xy$, we will use the product and chain rules. Similarly with the term
${y^2}$ which is actually ${\left( {some\,f\,of\,x} \right)^2}$
Product Rule can be expressed as : ${\left( {FS} \right)^\prime } = F'S + FS'$
$
2x\dfrac{{dx}}{{dx}} + \left( {\dfrac{d}{{dx}}\left( { + 2x} \right).y + \left( {2x}
\right)\dfrac{d}{{dx}}y} \right) - 2y\dfrac{{dy}}{{dx}} + 1 = 0 \\
2x + 2y + 2x\dfrac{{dy}}{{dx}} - 2y\dfrac{{dy}}{{dx}} + 1 = 0 \\
$
Now we will take out all the common factors.
\[\dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {2x + 2y + 1} \right)}}{{2x - 2y}}\]
At $\left( {1,2} \right)$,we get
\[
\dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {2(1) + 2(2) + 1} \right)}}{{2(1) - 2(2)}} \\
= \dfrac{{ - \left( {2 + 4 + 1} \right)}}{{2 - 4}} \\
= \dfrac{7}{2} \\
\]
So the slope of the tangent line at point $(1,2)$ is equal to $m = \dfrac{7}{2}$
Let's form the equation of the tangent line at point $(1,2)$ having slope $m = \dfrac{7}{2}$ by putting the $(1,2)\,as\,(y_1,y_2)$ in the slope point form $y - y_1 = m(x - x_1)$
$
y - 2 = \dfrac{7}{2}(x - 1) \\
2\left( {y - 2} \right) = 7(x - 1) \\
2y - 2 = 7x - 7 \\
0 = 7x - 2y - 5 \\
$
Therefore, the equation of tangent line to the curve at point $(1,2)$ is $0 = 7x - 2y - 5$.

Additional Information: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.

Note: While substituting the terms make sure you are taking into account the degrees and signs of the terms as well. While applying the power rule make sure you have considered the power with their respective signs. Remember that the derivative of ${x^3}$ is \[3{x^2}\] and the derivative of ${x^2}$ is $2x$. While applying the product rule, keep the first term as it is and differentiate the second term, then differentiate the first term and keep the second term as it is or vice versa.