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anonymous

  • 5 years ago

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

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  1. amistre64
    • 5 years ago
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    (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
    • 5 years ago
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    amistre64 is right

  3. amistre64
    • 5 years ago
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    yay!!

  4. anonymous
    • 5 years ago
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    lmao... YAY!!!

  5. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    the system has one solution x=4 y=2

  7. anonymous
    • 5 years ago
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    would you please explain? .....I've no clue how to do it... :(

  8. anonymous
    • 5 years ago
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    k

  9. anonymous
    • 5 years ago
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    \[y=\frac{1}{2}x, 2x+y=10\rightarrow y=\frac{1}{2}x, y=10-2x\] 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
    • 5 years ago
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    ohh...got it...thanks a lot.... :)

  11. anonymous
    • 5 years ago
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    no prob

  12. anonymous
    • 5 years ago
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    hey....1 thing....where is the 25 ?

  13. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    ohh.....can you please solve this once again ?

  15. anonymous
    • 5 years ago
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    yeah so are these the equations? \[25y=\frac{1}{2}x\] \[2x+y=10\]

  16. anonymous
    • 5 years ago
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    yeah.....

  17. anonymous
    • 5 years ago
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    \[25y=\frac{1}{2}x \rightarrow y=\frac{1}{50}x\] 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
    • 5 years ago
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    ohhh...how confusing :(.....thanks anyways.... :)

  19. anonymous
    • 5 years ago
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    yeah sorry about that

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