Generate Quiz 6973047E76478

(980-490t2) = 0-2(490)t=-980t v=d s/dt=d/dt

Considering t=½;

So: v=d s/dt at t=½ is:

v=d s/dt=-980(1/2)

v = -490 cm/sec

We can rewrite sin(x)cos(x) as 1/2 sin(2x)

Maximum value of sin(2x) is at x = 45° because sin 90° = 1 (max).

Therefore, maximum value of sin(x)cos(x) = (1/2)×1 = 1/2 or 0.5.

2cos(2x) d/dx(sin(2x)=cos(2x)xd/dx(2x)=2cos(2x)

The right response is b) cos(x).

Justification:

The standard derivative rule can be used to determine the derivative of the trigonometric function sin(x):

The derivative of sin(x) is cos(x).

This is due to the fact that a function’s derivative is its rate of change. Cos(x) is the rate at which sin(x) changes.

Consequently, b) cos(x) is the right response.

The rate at which f(x) changes in relation to ‘x’ is the right response (a). The rate at which the function f(x) changes in relation to the independent variable ‘x’ is known as the derivative of the function f(x) with respect to ‘x’.

The derivative of f(x) with respect to ‘x’ can be defined mathematically as follows:

f'(x) = lim[h→0] (f(x+h) – f(x)) / h

As the increment h gets closer to zero, this expression shows the limit of the difference quotient. It provides the function f(x)’s instantaneous rate of change with respect to the variable ‘x’.

Consequently, the rate of change of f(x) with respect to ‘x’ is the proper definition of the derivative of f(x) with respect to ‘x’.