Question

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

29. \(y = {e^x}\cos x + \sin x,\left( {0,1} \right)\)

Short answer

An equation of the tangent line to the curve at the point \(\left( {0,1} \right)\) is \(y = 2x + 1\).

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:

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

The derivative of the function at \(x = 0\) as shown below:

\(\begin{aligned}y'\left( 0 \right) &= {e^0}\left( {\cos 0 - \sin 0} \right) + \cos 0\\ &= 1\left( {1 - 0} \right) + 1\\ &= 1 + 1\\ &= 2\end{aligned}\)

The slope of the tangent line is 2.

Determine the equation of the tangent line at the point \(\left( {0,1} \right)\) as shown below:

\(\begin{aligned}{c}y - 1 &= 2\left( {x - 0} \right)\\y - 1 &= 2x\\y &= 2x + 1\end{aligned}\)

Thus, the equation of the tangent line to the curve at the point \(\left( {0,1} \right)\) is \(y = 2x + 1\).