Expert: Scott A Wilson - 11/11/2009

Question
my problems (set 3):
In Exercises 4 and 9, find dy/dx. Use your grapher to support your analysis if your are unsure of your answer.
4. y=(x)sec(x)
9. y=cot(x)/(1+cot(x))
11. Find and equation of the line tangent to the graph of y=sin(x)+3 at x=π. Answer in terms of π.
13. Find an equation of the line tangent to the graph of y=(x²)sin(x) at x=3. You must round your answer to 3 decimal places.
18. Show that the graphs of y=tan(x) and y=cot(x) have no horizontal tangents.

Answer
4. This is a product rule.  That's y = f(x)g(x) where f(x)=x and g()=secx.  
The derivatives are f'(x)=1 and g'(x)=secx•tanx.
The final formula for y' is y = fg' + fg'.

9. Here we have y=f(x)/g(x) where f=cot(x) and g=1 + cot(x), so y'=(gf'-fg')/g².

11. The function is y=sin(x)+3, so at x=π, sin(x)=0.
It is known thato y'=cos(x).  At the point x=π, this has the value -1.
Therefore all that is left is given the point (π,0) and the slope -1,
use the formula y-y0 = m(x-x0) where (x0,y0) is the point and m is he slope.

13. Since the function is y=x²sin(x), the derivative is a product rule with f(x)=x², g(x)=sin(x).
It is known that f'(x)=2x and g(x)=cos(x).  Using the product rule given in 4, the derivative can be found.  We are given x=3, so y=9sin(3), and m=y'(3).

18. If y=tan(x), y'=sec²(x), and it is known that sec²(x) is always at least 1.
If y=ctn(x), y'=-csc²(x), and since csc²(x) is always at least 1, -csc²(x) is always less than or equal to -1.  Since the derivative is the slope, it can be seen that the slope can't be 0 since it is not even in (-1,1).