(7)

Step 1:

The function is and .

Slope of the tangent line equation is derivative of the function.

.

Differentiate on each side with respect to .

.

Product rule of derivatives : .

Substitute in derivative function.

Slope of the tangent line equation is .

Step 2:

Find the point of tangency.

Substitute in the function.

Tangent point is .

Find the equation of tangent line.

Point slope form of the line equation : .

Substitute and in point slope form of line equation.

Tangent line equation is .

Solution :

Tangent line equation is .

(8)

The function is

Differentiate on each side with respect to .

.

Find the critical points.

A critical number of a function is a number in the domain of such that either or does not exist.

Find the critical point by graphing the derivative function.

From the above graph , it is clear that the critical points are and .

The range is .

\

The test intervals are , , , and .

Interval	Test Value	Sign of	Conclusion
		\ \	Increasing
		\ \	Decreasing
			Increasing
			Decreasing
			Increasing

Therefore the function is increasing on the intervals , and  .

And the function is decreasing on the intervals and  . \ \

Solution :

The function is increasing on the intervals , and  .

And the function is decreasing on the intervals and  .

.

General solution for trigonometric equation is .

Consider the values of for .

For

For

For

The critical points are .

\

The test intervals are and .

Interval	Test Value	Sign of	Conclusion
		\ \	Increasing
		\ \	Increasing

\

Therefore the function is increasing on the intervals and .

\

Solution :

The function is increasing on the intervals and .