Question

How do you find the equation of the tangent line to the curve $y = \sin \left( {\sin x} \right)$ at $\left( {\pi ,0} \right)$ ?

Answer 1

Hint: In this question, we have to find out the equation of the tangent line with the help of the given equation and the given point. First, differentiate the given equation to get the slope. Then, put the point $\left( {\pi ,0} \right)$ and the slope in the slope-point form of the equation. Simplify the equation and you will get your answer.

Formula used: To find the slope m of a curve at a particular point, we differentiate the equation of the curve. If the given curve is $y = f\left( x \right)$ we evaluate, \[\dfrac{{dy}}{{dx}}\] or $f'\left( x \right)$ and substitute the value of x to find the slope.
Point-slope form of equation:
\[y - {y_1} = m\left( {x - {x_1}} \right)\]

Complete step by step solution:
We have to find out the tangent line to the curve $y = \sin \left( {\sin x} \right)$ at the point $\left( {\pi ,0} \right)$ .
$ \Rightarrow y = \sin \left( {\sin x} \right)$ …. (given)
Differentiating both the sides with respect to,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {\sin x} \right) \times \cos x$ …. (Using chain rule)$x$
Finding the differentiation at the point $\left( {\pi ,0} \right)$ ,
$ \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = \pi }} = \cos \left( {\sin \pi } \right) \times \cos \pi $
We know that $\cos \pi = - 1,\sin \pi = 0$ . Putting these values in the equation,
$ \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = \pi }} = \cos \left( {0^\circ } \right) \times \left( { - 1} \right)$
Now, we know that $\cos 0^\circ = 1$ . Putting in the above equation,
$ \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = \pi }} = 1 \times \left( { - 1} \right)$
Hence, our slope, m $ = - 1$ .
Now, we have a point and a slope. We will use the point-slope form of the equation to find the required equation.
\[ \Rightarrow y - {y_1} = m\left( {x - {x_1}} \right)\]
Putting in all the values, we get,
\[ \Rightarrow y - 0 = - 1\left( {x - \pi } \right)\]
Simplifying the equation,
\[ \Rightarrow y = - x + \pi \]
Shifting the variables to the left-hand side,
\[ \Rightarrow y + x = \pi \]

Hence, the required equation of tangent to the given curve is \[y + x = \pi \] .

Note: Steps to determine the equation of a tangent to a curve:
Find the derivative using the rules of differentiation.
Substitute the x-coordinate and the y-coordinate of the given point into the derivative to calculate the slope of the tangent.
Substitute the slope of the tangent and the coordinates of the given point into an appropriate form of the straight- line equation, i.e., \[y - {y_1} = m\left( {x - {x_1}} \right)\] where m is the slope of the tangent and \[\left( {{x_1},{y_1}} \right)\] is the given point.