Question

The slope of the tangent to the curve ${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$ at the point $\left( {1,3} \right)$ is

Answer 1

Hint: Tangent is a line that touches the curve at only one point.
Use differentiation to find the slope of the tangent to the curve at the given point.
The given function is an implicit function, in which $x$ is dependent and $y$ is an independent variable, given in terms of both the variables $x$ and $y$.
For example: ${x^2} + 6xy + 2{y^2} = 0$ is an explicit function.
Thus, after differentiating the given function except for the derivative $\dfrac{{dy}}{{dx}}$ , try to bring all the other terms on the same side of the equation.
Remember that the derivative $\dfrac{{dy}}{{dx}}$ is the required slope of the tangent.

Complete step-by-step answer:
Given that: ${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$
On differentiating both sides with respect to $x$ .
Using differential: $\dfrac{{d{{\left( x \right)}^n}}}{{dx}} = n{x^{n - 1}}$
$2\left( {y - {x^5}} \right)\left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = {\left( {1 + {x^2}} \right)^2} + 2x\left( {1 + {x^2}} \right)\left( {2x} \right)$
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = \dfrac{{{{\left( {1 + {x^2}} \right)}^2} + 4{x^2}\left( {1 + {x^2}} \right)}}{{2\left( {y - {x^5}} \right)}}\]
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = \dfrac{{\left( {1 + {x^2}} \right)\left[ {\left( {1 + {x^2}} \right) + 4{x^2}} \right]}}{{2\left( {y - {x^5}} \right)}}\]
\[ \Rightarrow \left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = \dfrac{{\left( {1 + {x^2}} \right)\left( {1 + 5{x^2}} \right)}}{{2\left( {y - {x^5}} \right)}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 + {x^2}} \right)\left( {1 + 5{x^2}} \right)}}{{2\left( {y - {x^5}} \right)}} + 5{x^4}\]
Find the slope of the tangent:
The slope of the tangent to the curve $y = f\left( x \right)$ at the point $\left( {{x_0},{y_0}} \right)$ is given by \[{\left. {\dfrac{{dy}}{{dx}}} \right]_{\left( {{x_0},{y_0}} \right)}}\]
\[{\left. {\dfrac{{dy}}{{dx}}} \right]_{\left( {1,3} \right)}} = \dfrac{{\left( {1 + {1^2}} \right)\left( {1 + 5{{\left( 1 \right)}^2}} \right)}}{{2\left( {3 - {1^5}} \right)}} + 5{\left( 1 \right)^4}\]
$
   \Rightarrow \dfrac{{2 \times 6}}{{2 \times 2}} + 5 \\
   \Rightarrow 3 + 5 \\
   \Rightarrow 8 \\
 $

The slope of the tangent to the curve ${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$ at the point $\left( {1,3} \right)$ is 8.

Note: In a similar question equation of the tangent can be asked further. Use the equation of the straight line passing through a given point $\left( {\mathop x\nolimits_0 ,\mathop y\nolimits_0 } \right)$ having finite slope $m$:
                                            $y - \mathop y\nolimits_0 = m\left( {x - \mathop x\nolimits_0 } \right)$
Another entity, normal, is associated with the tangent.
The normal line to the curve at a given point is perpendicular to the tangent at that point.
Thus the slope of the normal line $ = - \dfrac{1}{{{\text{slope of tangent}}}}$
We know, the slope of the tangent line to the curve $y = f\left( x \right)$ is given by
                      The slope of the tangent line $ = \dfrac{{dy}}{{dx}}$ or $f'\left( x \right)$
Thus, the slope of the normal line to the curve $y = f\left( x \right)$ is given by
                     The slope of the normal line $ = - \dfrac{1}{{\dfrac{{dy}}{{dx}}}}{\text{ or }} - \dfrac{1}{{f'\left( x \right)}}$