The sign of the second derivative tells us what the concavity of the original function is.  If the second derivative is positive, then the function is concave up.  If the second derivative is negative, the function is concave down.  So if the second derivative is positive, then the function would be concave up.

We will solve for this by finding the higher order derivatives up until we reach the fourth order derivative.

We begin by finding the third order derivative.

And since the third order derivative is  it does not exist.  So this solution does not exist.

If  is decreasing, then its derivative is negative. The derivative of  is , so this is telling us that  is negative.

For  to be decreasing,  would have to be negative, which we don't know.

 being negative has nothing to do with its slope. 

For  to be decreasing, its derivative  would need to be negative, or, alternatively  would have to be concave down, which we don't know.

Thus, the only correct answer is that  is negative.

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see  hits zero at  and . Constructing an interval test,

 we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen  and  to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits  at . Thus, the intervals to test for the second derivative are 

.  Plugging in  and , we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

If this step is confusing, try drawing it out on a number line -- the first interval is from  to , the second from  to infinity. They only overlap on the smaller interval of  to .

Thus, our final answer is 

We see the answer is  when we use the product rule.

The  in the original expression is a constant and can be multiplied to the identity written above. 

When dealing with the derivative of , it is important to keep the standalone  in the denominator in absolute value bars and to make sure there is a  under the radical.

First, take the derivative of the function

Especially when given inverse trigonometry derivative questions, be on the lookout for multiple functions embedded in the same problem. For example, in this problem there is both an outer function () and an inner function (). Because there is more than one function, chain rule must be applied; thus, the derivative of the inner function must be multiplied to the derivative of the outer function. 

To reach the final answer, must be evaluated at .

This problem requires chain rule; there is an outer function () and an inner function ().

To simplify it further, square the  and multiply the .

This problem requires chain rule, because there is an outer and inner function.

The outer function is  and the inner is . 

To fully simplify the expression, multiply the constants and square the  term.