anonymous
  • anonymous
please find the derivative of y= sec squared x + tan squared X
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • 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 :)
anonymous
  • anonymous
amistre64 is right
amistre64
  • amistre64
yay!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
lmao... YAY!!!
anonymous
  • 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
anonymous
  • anonymous
the system has one solution x=4 y=2
anonymous
  • anonymous
would you please explain? .....I've no clue how to do it... :(
anonymous
  • anonymous
k
anonymous
  • 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.
anonymous
  • anonymous
ohh...got it...thanks a lot.... :)
anonymous
  • anonymous
no prob
anonymous
  • anonymous
hey....1 thing....where is the 25 ?
anonymous
  • 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... ?
anonymous
  • anonymous
ohh.....can you please solve this once again ?
anonymous
  • anonymous
yeah so are these the equations? \[25y=\frac{1}{2}x\] \[2x+y=10\]
anonymous
  • anonymous
yeah.....
anonymous
  • 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
anonymous
  • anonymous
ohhh...how confusing :(.....thanks anyways.... :)
anonymous
  • anonymous
yeah sorry about that

Looking for something else?

Not the answer you are looking for? Search for more explanations.