Question

please find the derivative of y= sec squared x + tan squared X

Answer 1

(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 :)

Answer 2

amistre64 is right

Answer 3

yay!!

Answer 4

lmao... YAY!!!

Answer 5

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

Answer 6

the system has one solution
x=4 y=2

Answer 7

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

Answer 8

k

Answer 9

\[y=\frac{1}{2}x, 2x+y=10\rightarrow y=\frac{1}{2}x, y=10-2x\]
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.

Answer 10

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

Answer 11

no prob

Answer 12

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

Answer 13

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... ?

Answer 14

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

Answer 15

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

Answer 16

yeah.....

Answer 17

\[25y=\frac{1}{2}x \rightarrow y=\frac{1}{50}x\]
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

Answer 18

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

Answer 19

yeah sorry about that