Question

27-36Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

34. \({x^2}{y^2} = {\left( {y + 1} \right)^2}\left( {4 - {y^2}} \right),\,\,\left( {2\sqrt 3 ,1} \right)\)

Short answer

The required equation of the tangent line is \(y = - \frac{{\sqrt 3 }}{5}x + \frac{{11}}{5}\).

Step-by-step solution

Write the definition of implicit differentiation, the formula of the derivatives of trigonometric functions and the product rule

Implicit differentiation: Differentiate the both sides of the equation with respect to \(x\) and then solve the obtained result for \(\frac{{dy}}{{dx}}\)this process is known as Implicit differentiation.

The product rule: If \(f\) and \(g\) are differentiable functions, then \(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\).

Theequation of the tangent line to the curve at the given point \(\left( {{x_1},{y_1}} \right)\)is \(y - {y_1} = \frac{{dy}}{{dx}}\left( {x - {x_1}} \right)\).

Find \(\frac{{dy}}{{dx}}\)by implicit differentiation

Consider the equation, \({x^2}{y^2} = {\left( {y + 1} \right)^2}\left( {4 - {y^2}} \right)\).

Differentiate both the sides of the equation w.r.t. \(x\)and simplify by using the formula of the product rule.

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^2}{y^2}} \right) &= \frac{d}{{dx}}{\left( {y + 1} \right)^2}\left( {4 - {y^2}} \right)\\{x^2} \cdot 2y\frac{{dy}}{{dx}} + {y^2} \cdot 2x &= {\left( {y + 1} \right)^2}\left( { - 2y\frac{{dy}}{{dx}}} \right) + \left( {4 - {y^2}} \right) \cdot 2\left( {y + 1} \right)\frac{{dy}}{{dx}}\\2{x^2}y\frac{{dy}}{{dx}} + 2x{y^2} &= - 2y\frac{{dy}}{{dx}}{\left( {y + 1} \right)^2} + 2\frac{{dy}}{{dx}}\left( {4 - {y^2}} \right)\left( {y + 1} \right)\\2x{y^2} &= \frac{{dy}}{{dx}}\left( {2\left( {4 - {y^2}} \right)\left( {y + 1} \right) - 2y{{\left( {y + 1} \right)}^2} - 2{x^2}y} \right)\\\frac{{dy}}{{dx}} &= \frac{{x{y^2}}}{{\left( {4 - {y^2}} \right)\left( {y + 1} \right) - y{{\left( {y + 1} \right)}^2} - {x^2}y}}\end{aligned}\)

Step 3: Find an equation of the tangent line to the curve at the given point, \(\left( {2\sqrt 3 ,1} \right)\)

Put \(x = 2\sqrt 3 \) and \(y = 1\)into \(\frac{{dy}}{{dx}} = \frac{{x{y^2}}}{{\left( {4 - {y^2}} \right)\left( {y + 1} \right) - y{{\left( {y + 1} \right)}^2} - {x^2}y}}\).

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{{x{y^2}}}{{\left( {4 - {y^2}} \right)\left( {y + 1} \right) - y{{\left( {y + 1} \right)}^2} - {x^2}y}}\\ &= \frac{{\left( {2\sqrt 3 } \right){{\left( 1 \right)}^2}}}{{\left( {4 - {{\left( 1 \right)}^2}} \right)\left( {1 + 1} \right) - \left( 1 \right){{\left( {1 + 1} \right)}^2} - {{\left( {2\sqrt 3 } \right)}^2}\left( 1 \right)}}\\ &= \frac{{2\sqrt 3 }}{{6 - 4 - 12}}\\ &= - \frac{{\sqrt 3 }}{5}\end{aligned}\)

Now put \(x = 2\sqrt 3 \), \(y = 1\) and \(\frac{{dy}}{{dx}} = 1\) into \(y - {y_1} = \frac{{dy}}{{dx}}\left( {x - {x_1}} \right)\).

\(\begin{aligned}y - 1 &= - \frac{{\sqrt 3 }}{5}\left( {x - 2\sqrt 3 } \right)\\y &= - \frac{{\sqrt 3 }}{5}x + \frac{6}{5} + 1\\y &= - \frac{{\sqrt 3 }}{5}x + \frac{{11}}{5}\end{aligned}\)

Thus, an equation of the tangent line to the curve at the given point, \(\left( {2\sqrt 3 ,1} \right)\) is \(y = - \frac{{\sqrt 3 }}{5}x + \frac{{11}}{5}\).