2.3-2.4 LT check

If f(x)=(x−1)^2sinx , then f'(0)=

1

A particle starts at time t=0 and moves along the x-axis so that its position at any time t ≥ 0 is given by x(t)=(t−1)^3(2t−3) . Find the value of t when the particle is moving and the acceleration is zero.

1 point is earned for derivative of their v from (a) or (b) 1 point is earned for solving for v'(t)=0 and solves for at least 2 roots 1 point is earned for answer a( t )= v ′( t ) =2( t−1 )( 8t−11 )+8( t−1 )^2 =6( t−1 )( 4t−5 ) a( t )=0 when t=1,t=5/4 but particle not moving at t=1 so t=5/4

What is the instantaneous rate of change at x=2 of the function f given by f(x)=(x^2−2)/(x−1) ?

2

If y=x2sin2x, then dy/dx=

2x(sin2x+xcos2x)

If y=sin(3x), then dy/dx=

3cos(3x)

If y=sin3x, then dy/dx=

3sin^2xcosx

(Table) The table above gives values of the differentiable functions f and g and their derivatives at x = 1. If h(x) = (2 f (x) + 3)(1 + g(x)), then h'(1) =

44

If y=5x√x^2+1, then dy/dx at x=3 is

45 /√10+5√10

d/dx(sin^3(x^2))=

6xsin^2(x^2)cos(x^2)

If f(x)=tan(2x) , then f′(π6)=

8

A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by v(t) = 1−sin(2πt) . Find the acceleration a(t) of the particle at any time t.

a(t)=v'(t) =−2πcos(2πt)

Let g be the function defined by g(x) = xf(x). Find an equation for the line tangent to the graph of g at x=2.

g'(x) = f(x)+xf'(x) g'(2) = f(2)+2f'(2) = 6+2(-1)=4 g(2)=2f(2)=12 Tangent line is y=4(x-2)+12

(Table) The function h is defined by h(x)=ln(f(x)) . Find h'(3). Show the computations that lead to your answer.

h'(x)= f'(x)/f(x) then solve for 3

An equation of the line tangent to the graph of f(x)=x(1−2x)^3 at the point(1,−1) is

y=−7x+6

If f(x)=cos^3(4x) , then f'(x)=

−12cos^2(4x)sin(4x)

The graph of the function f, consisting of two line segments, is shown in the figure above. Let g be the function given by g (x)=2x+1, and let h be the function given by h (x)=f (g (x)). h′ (1) ?

-4

If f(x)=(x^2+3x+2)/(x+3), then f′(x)=

x^2+6x+7/(x+3)^2