# Find an equation of the tangent line to the curve at the given point. y = x , (81, 9)

The aim of this question is to deduce the equation of a tangent line of a curve at any point on the curve.

For any given function $y = f(x)$, the equation of its tangent line is defined by the following equation:

$\boldsymbol{ y – y_1 = \frac{ dy }{ dx } ( x – x_1 ) }$

Here, $( x_1 , y_1 )$ is the point on the curve $y = f(x)$ where the tangent line is to be evaluated and $\dfrac{ dy }{ dx }$ is the value of the derivative of the  subject curve evaluated at the required point.

Given that:

$y = \sqrt{ x }$

Calculating the derivative of $y$ with respect to $x$:

$\frac{ dy }{ dx } = \frac{ 1 }{ 2 \sqrt{ x } }$

Evaluating above derivative at given point $( 81 , 9 )$:

$\frac{ dy }{ dx } |_{ ( 81 , 9 ) } = \frac{ 1 }{ 2 \sqrt{ 81 } }$

$\frac{ dy }{ dx } |_{ ( 81 , 9 ) } = \frac{ 1 }{ 2 ( 9 ) }$

$\frac{ dy }{ dx } |_{ ( 81 , 9 ) } = \frac{ 1 }{ 18 }$

The equation of a tangent line with slope $\dfrac{ dy }{ dx }$ and point $( x_1 , y_1 )$ is defined as:

$y – y_1 = \frac{ dy }{ dx } ( x – x_1 )$

Substituting values of $\dfrac{ dy }{ dx } = \dfrac{ 1 }{ 18 }$ and point $( x_1 , y_1 ) = ( 81 , 9 )$ in above equation:

$y – 9 = \frac{ 1 }{ 18 } ( x – 81 )$

$y – 9 = \frac{ 1 }{ 18 } x – \frac{ 1 }{ 18 } 81$

$y – 9 = \frac{ 1 }{ 18 } x – \frac{ 9 }{ 2 }$

$y = \frac{ 1 }{ 18 } x – \frac{ 9 }{ 2 } + 9$

$y = \frac{ 1 }{ 18 } x + \frac{ – 9 + ( 2 ) ( 9 ) }{ 2 }$

$y = \frac{ 1 }{ 18 } x + \frac{ – 9 + 18 }{ 2 }$

$\boldsymbol{ y = \frac{ 1 }{ 18 } x + \frac{ 9 }{ 2 } }$

## Numerical Result

$\boldsymbol{ y = \frac{ 1 }{ 18 } x + \frac{ 9 }{ 2 } }$

## Example

Find an equation of the tangent line to the curve $y = x$ at $(1, 10)$.

Here:

$\frac{ dy }{ dx } = 1$

Using the tangent equation with $\dfrac{ dy }{ dx } = 1$ and point $( x_1 , y_1 ) = ( 1 , 10 )$:

$y – y_1 = \frac{ dy }{ dx } ( x – x_1 )$

$y – 10 = ( 1 ) ( x – 1 )$

$y = ( 1 ) ( x – 1 ) + 10 = x – 1 + 10$

$\boldsymbol{ y = x + 9 }$