Question

27-30: Find an equation of the tangent line to the curve at the given point.

28. \(y = x + \sin x,\left( {\pi ,\pi } \right)\)

Short answer

An equation of the tangent line to the curve at the point \(\left( {\pi ,\pi } \right)\) is \(y = \pi \).

Step-by-step solution

Derivative of a trigonometric function

The derivative of a trigonometric function is shown below:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\csc x} \right) &= - \csc x\cot x\\\frac{d}{{dx}}\left( {\cos x} \right) &= - \sin x\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\sec x} \right) &= \sec x\tan x\\\frac{d}{{dx}}\left( {\tan x} \right) &= {\sec ^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\cot x} \right) &= - {\csc ^2}x\end{aligned}\)

Determine the equation of the tangent line to the curve

Evaluate the derivative of the function as shown below:

\(\begin{aligned}y'\left( x \right) &= \frac{d}{{dx}}\left( {x + \sin x} \right)\\ &= \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {\sin x} \right)\\ &= 1 + \cos x\end{aligned}\)

The derivative of the function at \(x = \pi \) as shown below:

\(\begin{aligned}y'\left( \pi \right) &= 1 + \cos \pi \\ &= 1 + \left( { - 1} \right)\\ &= 0\end{aligned}\)

The slope of the tangent line is 0.

Determine the equation of the tangent line at the point \(\left( {\pi ,\pi } \right)\) as shown below:

\(\begin{aligned}y - \pi &= 0\left( {x - \pi } \right)\\y - \pi &= 0\\y &= \pi \end{aligned}\)

Thus, the equation of the tangent line to the curve at the point \(\left( {\pi ,\pi } \right)\) is \(y = \pi \).