Question

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

Answer 1

Hint: We are asked to find the equation of tangent line for the given curve, for this we will first find the slope of the tangent line by differentiating the given curve by using the differentiating identity i..e., \[\dfrac{d}{{dx}}\sin x = \cos x\] and to find the tangent line equation we will use the slope-point formula which is given by, \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\], where m is the slope and \[\left( {{x_1},{y_1}} \right)\] is the point, and finally substituting the given point and the slope we will get the required tangent equation.

Complete step-by-step solution:
Given equation of curve is \[y = \sin \left( {\sin x} \right)\],
We have to find the equation of the tangent at the point \[\left( {\pi ,0} \right)\],
Now differentiating both sides we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\sin \left( {\sin x} \right)\],
Now applying derivatives i.e.,\[\dfrac{d}{{dx}}\sin x = \cos x\] we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {\sin x} \right)\dfrac{d}{{dx}}\sin x\],
Now differentiating the inner part, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {\sin x} \right)\left( {\cos x} \right)\],
Now we know that slope is given by \[\dfrac{{dy}}{{dx}}\] at the given point, here the point is \[\pi \],
So, slope=\[{\dfrac{{dy}}{{dx}}_{x = \pi }} = \cos \left( {\sin \pi } \right)\left( {\cos \pi } \right)\],
Now using the trigonometric ratios i.e, \[\cos \pi = - 1\], we get,
Slope=\[{\dfrac{{dy}}{{dx}}_{x = \pi }} = \cos \left( 0 \right)\left( { - 1} \right)\],
Now we know that \[\cos 0 = 1\], then the slope will be,
Slope=\[{\dfrac{{dy}}{{dx}}_{x = \pi }} = 1\left( { - 1} \right)\],
Now multiplying we get,
Slope=\[{\dfrac{{dy}}{{dx}}_{x = \pi }} = - 1\],
Now we have to find the equation of tangent using the slope point formula which is given by, \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\], where m is the slope and \[\left( {{x_1},{y_1}} \right)\] is the point,
Now substituting the values in the formula \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\] and here \[\left( {{x_1},{y_1}} \right) = \left( {\pi ,0} \right)\] and \[m = - 1\],
So, the equation of tangent becomes,
\[ \Rightarrow \left( {y - 0} \right) = - 1\left( {x - \pi } \right)\],
Now simplifying we get,
\[ \Rightarrow y = - x + \pi \],
So the equation of the tangent line is \[y = \pi - x\].

\[\therefore \]The tangent of the equation of tangent line to the curve \[y = \sin \left( {\sin x} \right)\] at \[\left( {\pi ,0} \right)\] is \[y = \pi - x\].

Note: In order to find the equation of a tangent, we:
Differentiate the equation of the curve
Substitute the \[x\] value into the differentiated equation to find the gradient
Substitute the \[x\] value into the original equation of the curve to find the \[y\]-coordinate
Substitute your point on the line and the gradient into \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\].