Question

Can someone help walk me through how to find the derivative of tan(x) using the limit definition?

Answer 1

\[\lim_{h \rightarrow 0} \frac{ (tanx+h)-\tan(x) }{ h }\]

Answer 2

go ahead.
\[ \large (\tan x)'=\lim_{h\to0}\frac{\tan(x+h)-\tan x}{h} \]

Answer 3

check your parenthesis.

Answer 4

\[\lim_{h \rightarrow }(\frac{ tanx+\tanh }{ 1-tanxtanh }-tanx)/h\]

Answer 5

oh woops, that's what I meant.

Answer 6

but then I have no idea how to simplify that.

Answer 7

\[ \large \tan(x+h)=? \]

Answer 8

recall a trig identity for tan (x+h)...

Answer 9

\[\frac{ tanx+\tanh }{ 1-tanxtanh}\]

Answer 10

great

Answer 11

yeah

Answer 12

that's where I'm stuck

Answer 13

so we have
\[ \large (\tan x)'=\lim_{h\to0}\frac{\frac{\tan x+\tan h}{1-\tan x\tan h}-\tan x}{h} \]

Answer 14

yep, that's where I got

Answer 15

I think I maybe got it

Answer 16

add the fractions in the numerator.

Answer 17

convert than tans to sin/cos then factor out the sinx/cosx to get all the trig h's in the limit...

Answer 18

nevermind, that didn't work

Answer 19

ok

Answer 20

so
\[ \large (\tan x)'=\frac{\tan x+\tan h-\tan x+\tan^2x\tan h}{h(1-\tan x\tan h)} \]

Answer 21

right?

Answer 22

yes.

Answer 23

simplify and factor.

Answer 24

what am I factoring out? tanh?

Answer 25

yes

Answer 26

ok.

Answer 27

\[\lim_{h \rightarrow 0}\frac{ \tanh(1-\tan ^{2}x) }{ (1_tanxtanh)(h) }\]

Answer 28

sorry, the denominator should be 1-tanxtanh

Answer 29

times h

Answer 30

no

Answer 31

now I'm lost again lol

Answer 32

\[ \large (\tan x)=\lim_{h\to0}\frac{\tan h(1+\tan^2x)}{h(1-\tan x\tan h)} \]

Answer 33

\[\lim_{h \rightarrow 0}\frac{ (\tanh)(1-\tan ^{2}x) }{ (h)(1-tanxtanh) }\]

Answer 34

that's what I meant, sorry. I have trouble with the equation maker

Answer 35

the subtraction sign in the numerator should be an addition @nasryn

Answer 36

oh.

Answer 37

i know. it is better to use the keyboard directly (if u know latex).

Answer 38

now. let's look at the pieces:
\[ \large \lim_{h\to0}\frac{\tan h}{h}=? \]

Answer 39

the only thing I can think of to do next is to make tanh = sinh/cosh, then cancel the sinh with sinh/h which the liimit of that would be 1

Answer 40

yes.

Answer 41

I don't know that identity for tanh/h directly

Answer 42

\[ \large \lim_{h\to0}\frac{\tan h}{h}=\lim_{h\to0}\frac{\sin h}{h}\cos h=1\cdot 1=1 \]
there is no identity for this limit.

Answer 43

oh! wow! I feel silly

Answer 44

finally
\[ \large \lim_{h\to0}\frac{1}{1-\tan x\tan h}=? \]

Answer 45

wait.

Answer 46

would tanh/h be (sinh)/(h)(cosh)

Answer 47

yes. that's what i wrote.
\[ \large \frac{\tan h}{h}=\frac{\sin h}{\cos h}\frac{1}{h} \]

Answer 48

oh ok

Answer 49

sorry. i had a typo.

Answer 50

so the answer is just 1+tan^2x?

Answer 51

ohh and that's sec^2(x)!

Answer 52

correct.

Answer 53

yes

Answer 54

:) thanks!

Answer 55

but now how do I get rid of tanxtanh in the denominator?

Answer 56

oh, just plug in. :P

Answer 57

\[ \large \lim_{h\to0}\frac{1}{1-\tan x\tan h}=\frac{1}{1-\tan x\tan0}=1 \]

Answer 58

that was rough, I hope that's not on my exam. :P

Answer 59

so
\[ \large \lim_{h\to0}\frac{\tan h(1+\tan^2x)}{h(1-\tan x\tan h)}=1+\tan^2x=\sec^2x \]

Answer 60

the best of luck in your test.

Answer 61

thanks.