let ( f ) be the function defined by ( f(x)=cos (2 x)+e^{sin x} ).\nlet ( g ) be a differentiable function…

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

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

Answer

(a)

Explanation:

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)

Answer:

The slope of the tangent line is (-1)

(b)

Explanation:

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)

Answer:

(k^{\prime}(\pi)=-1)

(c)

Explanation:

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:

(m^{\prime}(2)=3)