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anonymous
 one year ago
derivative of tan x using first principle
anonymous
 one year ago
derivative of tan x using first principle

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dunno if it helps to use the sin/cos definintion

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i can never recall the tan(a+b) formula .. can you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i used quotient rule and got sec^2 x which i know is right but the first principle rule is troubling

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433521302532:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1wondering 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{sin(x)}{cos(x+h)}\frac{cos(h) }{h}+\frac{cos(x)}{cos(x+h)}\frac{ sin(h) }{h}\frac{sin(x)}{h~cos(x)}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1do you recall what the cos(h) and sin(h) parts limit to?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i need to follow in similar manner

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\frac{sin(x+h)}{cos(x+h)}\frac{sin(x)}{cos(x)}}{h}\] \[\frac{cos(x)sin(x+h)sin(x)cos(x+h)}{h~cos(x)cos(x+h)}\] \[\frac{cos(x)}{cos(x)cos(x+h)} \frac{sin(x)sin(h)+cos(x)sin(h)}{h} \\ \frac{sin(x)}{cos(x)cos(x+h)}\frac{cos(x)cos(h)sin(x)sin(h)}{h}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1got a typo ... ***sin(x)cos(h)*** in the expansion of sin(x+h)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i cant see an error other than that .... but im biased. how is your tan version coming along?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if there is any mistake please point it out

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1i recall that sin(h)/h limits to 1 is there a similar 'known' for cos(h)/h ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you ill try and correct the mistake :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1[cos(h)1]/h is what we need to deal with the cosine.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1for 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
 one year ago
Best ResponseYou've already chosen the best response.1so im sure theres some more factoring to do in order to approach our goal
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