anonymous
  • anonymous
derivative of tan x using first principle
Mathematics
jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
dunno if it helps to use the sin/cos definintion
amistre64
  • amistre64
i can never recall the tan(a+b) formula .. can you?
anonymous
  • anonymous
i used quotient rule and got sec^2 x which i know is right but the first principle rule is troubling

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anonymous
  • anonymous
|dw:1433521302532:dw|
amistre64
  • amistre64
wondering if this works any simpler | sin(x)cos(h) + sin(h)cos(x) sin(x) | 1 | ---------------------- - ----- | --- | cos(x)cos(h) - sin(x)sin(h) cos(x) | h
amistre64
  • amistre64
\[\frac{sin(x)}{cos(x+h)}\frac{cos(h) }{h}+\frac{cos(x)}{cos(x+h)}\frac{ sin(h) }{h}-\frac{sin(x)}{h~cos(x)}\]
amistre64
  • amistre64
do you recall what the cos(h) and sin(h) parts limit to?
anonymous
  • anonymous
anonymous
  • anonymous
i need to follow in similar manner
amistre64
  • amistre64
i though you needed to work it by first principles so far ive got it to sin(x)cos(x) cos(h) + cos^2(x)sin(h) -sin(x)[cos(x)cos(h) - sin(x)sin(h)] ----------------------------------------------------------- h(cos(x+h)cos(x)) which allows us to get the h parts lined up
anonymous
  • anonymous
then apply limits
amistre64
  • amistre64
\[\frac{\frac{sin(x+h)}{cos(x+h)}-\frac{sin(x)}{cos(x)}}{h}\] \[\frac{cos(x)sin(x+h)-sin(x)cos(x+h)}{h~cos(x)cos(x+h)}\] \[\frac{cos(x)}{cos(x)cos(x+h)} \frac{sin(x)sin(h)+cos(x)sin(h)}{h} \\ -\frac{sin(x)}{cos(x)cos(x+h)}\frac{cos(x)cos(h)-sin(x)sin(h)}{h}\]
amistre64
  • amistre64
got a typo ... ***sin(x)cos(h)*** in the expansion of sin(x+h)
amistre64
  • amistre64
i cant see an error other than that .... but im biased. how is your tan version coming along?
anonymous
  • anonymous
anonymous
  • anonymous
if there is any mistake please point it out
amistre64
  • amistre64
i dont see that you dropped the denominator of the top, down to the bottom the GCD of cos(x) and cos(x+h), is cos(x)cos(x+h)
amistre64
  • amistre64
i recall that sin(h)/h limits to 1 is there a similar 'known' for cos(h)/h ?
anonymous
  • anonymous
thank you ill try and correct the mistake :)
amistre64
  • amistre64
[cos(h)-1]/h is what we need to deal with the cosine.
amistre64
  • amistre64
for some recollection ... f' of sin(x) sin(x+h) - sin(x) ----------------- h sin(x)cos(h) +cos(x)sin(h) - sin(x) ----------------------------------- h sin(x)[cos(h)-1] +cos(x)sin(h) ------------------------------- h sin(x)[cos(h)-1]/h +cos(x)sin(h)/h sin(x)(0) +cos(x)(1) cos(x)
amistre64
  • amistre64
so im sure theres some more factoring to do in order to approach our goal

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