Question

57-60 Find an equation of the tangent line to the curve at the given point.

59. \(y = {\bf{sin}}\left( {{\bf{sin}}\,x} \right)\), \(\left( {\pi ,{\bf{0}}} \right)\)

Short answer

The equation of the tangent line is \(y = - x + \pi \).

Step-by-step solution

Step 1:Find the value of \(y'\)

Find the value of \(y'\) for the equation\(y = \sin \left( {\sin x} \right)\).

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}x}}& = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\sin \left( {\sin x} \right)} \right)\\ &= \cos \left( {\sin x} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\sin x} \right)\\ &= \cos \left( {\sin x} \right)\left( {\cos x} \right)\end{aligned}\)

Find the slope of the tangent at point \(\left( {\pi ,{\bf{0}}} \right)\)

The slope is shown below:

\(\begin{aligned}{\left. {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right|_{\left( {\pi ,0} \right)}} &= \cos \left( {\sin \pi } \right)\cos \left( \pi \right)\\ &= \cos \left( 0 \right)\left( { - 1} \right)\\ &= - 1\end{aligned}\)

Find the equation forthe tangent

The equation of tangentpassing through \(\left( {\pi ,0} \right)\) with slope \( - 1\) is:

\(\begin{array}{c}y - 0 = - 1\left( {x - \pi } \right)\\y = - x + \pi \\y = - x + \pi \end{array}\)

So, the equation of tangent is \(y = - x + \pi \).