Exam 3: Differentiation Rules

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x f(x) (x) g(x) (x) -6 7 -8 -6 7 -4 1 -5 0 5 -2 3 -2 4 3 0 5 0 6 1 2 -5 1 6 -1 4 -3 3 4 -3 6 1 5 0 -5 -Find ddx(f(x)g(x)) when x=2\frac { d } { d x } ( f ( x ) \cdot g ( x ) ) \text { when } x = - 2

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The displacement of a particle is given by x=0.03sin(20πt+π2) meters x = 0.03 \sin \left( 20 \pi t + \frac { \pi } { 2 } \right) \text { meters } . Find all times t > 0 where (a) The displacement attains its maximum value.(b) The velocity attains its maximum value.(c) The acceleration attains its maximum value.

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Below is a table containing information about the differentiable functions f and g. x -2 -1 0 1 2 f(x) 4 3 5 2 1 (x) 1 4 2 5 3 g(x) 5 2 1 3 4 (x) 3 5 4 1 2 Suppose also that F(x)=f(x)+g(x)F ( x ) = f ( x ) + g ( x ) , G(x)=f(x)g(x)G ( x ) = f ( x ) \cdot g ( x ) , and H(x)=g(x)f(x)H ( x ) = \frac { g ( x ) } { f ( x ) } .(a) Find F(0)F ^ { \prime } ( 0 ) (b) Find G(2)G ^ { \prime } ( - 2 ) (c) Find H(2)H ^ { \prime } ( 2 )

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If f(x)=x43xf ( x ) = x ^ { 4 } - 3 x , find f(1)f ^ { \prime \prime } ( 1 ) .

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If the total cost for producing x units of a particular product is given by C(x), then the average cost of production those x units is given by A(x)=C(x)xA ( x ) = \frac { C ( x ) } { x } (a) If our cost function is C(x)=15,000+100x+2r3/2C ( x ) = 15,000 + 100 x + 2 r ^ { 3 / 2 } , what value of x will result in the minimum average cost? (b) What is the minimum average cost?

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Simplify the expression cos(2sin1x)\cos \left( 2 \sin ^ { - 1 } x \right) .

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The position function for a particle moving along the x-axis at time t > 0 (in seconds) has its x-coordinate given by x(t)=t48t3+6t2+40t20x ( t ) = t ^ { 4 } - 8 t ^ { 3 } + 6 t ^ { 2 } + 40 t - 20 .(a) Find a formula for the velocity and acceleration of the particle at any time t.(b) What is the velocity of the particle when t = 3? What does this value indicate?

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x f(x) (x) g(x) (x) -6 7 -8 -6 7 -4 1 -5 0 5 -2 3 -2 4 3 0 5 0 6 1 2 -5 1 6 -1 4 -3 3 4 -3 6 1 5 0 -5 -Find ddx(exg(x)) when x=0\frac { d } { d x } \left( e ^ { x } \cdot g ( x ) \right) \text { when } x = 0

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Consider the curve given by x=2sint+5,y=45cost,0t2πx = 2 \sin t + 5 , y = 4 - 5 \cos t , 0 \leq t \leq 2 \pi Find dydx\frac { d y } { d x } at the point corresponding to t=5π4t = \frac { 5 \pi } { 4 }

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If f(x)=sin2x, find f(π4)f ( x ) = \sin ^ { 2 } x , \text { find } f ^ { \prime } \left( \frac { \pi } { 4 } \right)

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The mass of a rod varies in such a way that the mass of a piece x meters long, measured from the left end, is x2x ^ { 2 } kilograms. Find the density in kg/m at the point 2 meters from the left end.

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If f(x)=exg(x), and if g(0)=3 and g(0)=2, find f(0)f ( x ) = e ^ { x } g ( x ) \text {, and if } g ( 0 ) = 3 \text { and } g ^ { \prime } ( 0 ) = - 2 \text {, find } f ^ { \prime } ( 0 ) \text {. }

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Find dydx\frac { d y } { d x } , (a) y=x25xy = \sqrt { x ^ { 2 } - 5 x } (b) y=sin(xex)x3+xy = \frac { \sin \left( x e ^ { x } \right) } { x ^ { 3 } + x } (c) y=x2exy = x ^ { 2 } \cdot e ^ { \sqrt { x } } (d) y=ex2+1y = e ^ { \sqrt { x ^ { 2 } + 1 } }

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If xy2y2=x3yx y - 2 y ^ { 2 } = x ^ { 3 } y find the value of dydx\frac { d y } { d x } at the point ( 2,3- 2,3 ).

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Find an equation of the tangent line to the graph of y=xlnxy = x \ln x at the point (1, 0).

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If xey=yx2x e ^ { y } = y x ^ { 2 } , find an expression for dydx\frac { d y } { d x } .

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Let y = x2x ^ { 2 } , x = 2, and Δ\Delta x = 1. Find the value of the differential dy.

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Find dydx\frac { d y } { d x } for the parametric curve given by x=t4t2+t,y=t3x = t ^ { 4 } - t ^ { 2 } + t , y = \sqrt [ 3 ] { t }

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At how many different values of x does the curve y = x 33 - 2x have a tangent line parallel to the line y=xy = x ?

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x f(x) (x) g(x) (x) -6 7 -8 -6 7 -4 1 -5 0 5 -2 3 -2 4 3 0 5 0 6 1 2 -5 1 6 -1 4 -3 3 4 -3 6 1 5 0 -5 -Find ddx(g(x)f(x)) when x=2\frac { d } { d x } \left( \frac { g ( x ) } { f ( x ) } \right) \text { when } x = - 2

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