anonymous 5 years ago please find the derivative of y= sec squared x + tan squared X

1. amistre64

(d/dx)sec(x)^2 + (d/dx)tan(x)^2: Let me think: u=sec(x) -> sec(x)tan(x); and, v=tan(x) -> sec^2(x) 2u*sec(x)tan(x) + 2v*sec^2(x) 2sec(x)sec(x)tan(x) + 2tan(x)sec(x)sec(x) 2sec^2(x)tan(x) + 2sec^2(x)tan(x) = 4sec^2(x)tan(x) If I did it correctly, this should be it :)

2. anonymous

amistre64 is right

3. amistre64

yay!!

4. anonymous

lmao... YAY!!!

5. anonymous

Thank you all..... :) can you please help in this problem too : I need to determine whether a system has no solution, one solution or infinitely solutions. 25. y=1/2x; 2x+y=10

6. anonymous

the system has one solution x=4 y=2

7. anonymous

would you please explain? .....I've no clue how to do it... :(

8. anonymous

k

9. anonymous

$y=\frac{1}{2}x, 2x+y=10\rightarrow y=\frac{1}{2}x, y=10-2x$ now substitute y= 1/2x into the second equation$\frac{1}{2}x=10-2x \rightarrow \frac{5}{2}x=10\rightarrow x=\frac{20}{5}\rightarrow x=4$ plug this value of x into the first equation to get the y value$y=\frac{1}{2}x, x=4\rightarrow y=\frac{1}{2}*4\rightarrow y=\frac{4}{2}= 2$ Another thing to notice is that y=1/2x has a positive slope while y=10-2x has a negative slope, which means they will intersect once. If both your y's have the same slope but different x and y intercepts they will never intersect because they are parallel. If they have the same slope and the same x and y intercepts, you will have infinetely many solutions.

10. anonymous

ohh...got it...thanks a lot.... :)

11. anonymous

no prob

12. anonymous

hey....1 thing....where is the 25 ?

13. anonymous

lol.... wait is the first equation 25y=1/2x bc you have 25. y=1/2x I just thought 25 was the number of the problem... ?

14. anonymous

ohh.....can you please solve this once again ?

15. anonymous

yeah so are these the equations? $25y=\frac{1}{2}x$ $2x+y=10$

16. anonymous

yeah.....

17. anonymous

$25y=\frac{1}{2}x \rightarrow y=\frac{1}{50}x$ plug this value for y into the second equation$y=\frac{1}{50}x, 2x+y=10\rightarrow 2x+\frac{1}{50}x=10$ $\frac{1}{50}x+\frac{100}{50}x=10\rightarrow \frac{101}{50}x=10$solve for x by finding a common denominator $x=\frac{500}{101}$ Plug this value in for x in the fist equation$y=\frac{1}{50}x, x= \frac{500}{101}\rightarrow y=\frac{1}{50}\frac{500}{101}=\frac{10}{101}$ These are your unique solutions

18. anonymous

ohhh...how confusing :(.....thanks anyways.... :)

19. anonymous