Question

let ( f ) be the function defined by ( f(x)=cos (2 x)+e^{sin x} ).
let ( g ) be a differentiable function. the table above gives values of ( g ) and its derivative ( g^{prime} ) at selected value
of ( x ).
let ( h ) be the function whose graph, consisting of five line segments, is shown in the figure above.
(a) find the slope of the line tangent to the graph of ( f ) at ( x=pi ).
(b) let ( k ) be the function defined by ( k(x)=h(f(x)) ). find ( k^{prime}(pi) ).
(c) let ( m ) be the function defined by ( m(x)=g(-2 x) cdot h(x) ). find ( m^{prime}(2) ).

Explanation:

(a)

Step1: Differentiate \(f(x)\)

Use the chain rule. The derivative of \(\cos(2x)\) is \(- 2\sin(2x)\), and the derivative of \(e^{\sin x}\) is \(e^{\sin x}\cos x\). So \(f^{\prime}(x)=-2\sin(2x)+e^{\sin x}\cos x\)

Step2: Substitute \(x = \pi\)

When \(x=\pi\), \(\sin(2\pi)=0\), \(\sin(\pi) = 0\), and \(\cos(\pi)=- 1\). Then \(f^{\prime}(\pi)=-2\sin(2\pi)+e^{\sin\pi}\cos\pi=0 + e^{0}\times(-1)=-1\)

Step1: Use the chain rule

By the chain rule \(k^{\prime}(x)=h^{\prime}(f(x))\cdot f^{\prime}(x)\). First, we know from part (a) that \(f^{\prime}(\pi)=-1\), and \(f(\pi)=\cos(2\pi)+e^{\sin\pi}=1 + 1=2\)

Step2: Find \(h^{\prime}(2)\)

Looking at the graph of \(h\), the slope of the line segment when \(x = 2\) (using the two - point formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\), for the segment passing through \((1,0)\) and \((3,-1)\) the slope \(h^{\prime}(x)\) (since it's a line segment, the derivative is the slope of the segment) \(h^{\prime}(2)=1\) (using the formula for the slope of the line segment from \((1,0)\) to \((3,2)\) (counting grid - points: \(\frac{2-0}{3 - 1}=1\))
Then \(k^{\prime}(\pi)=h^{\prime}(f(\pi))\cdot f^{\prime}(\pi)=h^{\prime}(2)\cdot(-1)=1\times(-1)=-1\)

Step1: Use the product rule

The product rule states that if \(m(x)=u(x)\cdot v(x)\) where \(u(x)=g(-2x)\) and \(v(x)=h(x)\), then \(m^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)\)
First, find \(u^{\prime}(x)\) using the chain rule. If \(u(x)=g(-2x)\), then \(u^{\prime}(x)=g^{\prime}(-2x)\times(-2)\)
When \(x = 2\), \(u(2)=g(-4)\), \(u^{\prime}(2)=-2g^{\prime}(-4)\), and \(v(2)=h(2)\), \(v^{\prime}(2)=h^{\prime}(2)\)
From the table \(g(-4) = 5\) and \(g^{\prime}(-4)=-1\), from the graph \(h(2)=-1\) and \(h^{\prime}(2)=1\)
\(u^{\prime}(2)=-2g^{\prime}(-4)=(-2)\times(-1) = 2\)

Step2: Calculate \(m^{\prime}(2)\)

\(m^{\prime}(2)=u^{\prime}(2)v(2)+u(2)v^{\prime}(2)=2\times(-1)+5\times1=3\)

Answer:

The slope of the tangent line is \(-1\)

(b)