Question

Diff the following functions,
1.) tan^-1((a+x)/(1-ax))
2.)tan^-1((a+btanx)/(b-atanx))
My question is can you substitute for constants? like a=tan(alpha)

Answer 1

HI!!

Answer 2

no i don't think so

Answer 3

oh

Answer 4

i think the idea of this question is this:
once you find the derivative of the first one, the derivative of the second will be easy by the chain rule

Answer 5

These questions are to be done by trigonometric subtitutions but I'm just stuck on the 2nd one, and I just found I did the first one wrong because you can't sub for constants according to you

Answer 6

hmm i may be wrong about this, i have never heard of a trig sub for a derivative, only for an integral
a quick wolfram check tells me the derivative of the first one is identical to the derivative of \(\tan^{-1}(x)\)

Answer 7

it is not at all clear to me why however
if you do it the long way you get \(\frac{1}{1+x^2}\)

Answer 8

Oh?
Here's an example,
y=sin^-1(root(1-x^2))
x=cos(theta)
theta=cos^-1(x)
y=sin^-1(root(1-cos^2(theta)))
y=sin^-1(root(sin^2(theta)))
y=sin^-1(sin(theta))
y=theta
y=cos^-1(x)
dy/dx= -1/root(1-x^2)

Answer 9

i see

Answer 10

Sometimes in inverse trig functions, you can sub and then derivative, it's en exercise in my textbook

Answer 11

heres a reference of derivatives for inverse trig.
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Derivatives_of_inverse_trigonometric_functions

then use chain rule

Answer 12

@dumbcow this is somehow different

Answer 13

i mean you can do it they way you suggested of course, but there is some trick involved using a trig sub

Answer 14

That is good and all Professor, but these questions are to be done with sub, doing the direct way would take time

Answer 15

oh i see... i think the trig sub just allows you to derive the formula for derivative

Answer 16

so would the sub be \(x=\tan(\theta)\)?

Answer 17

Yeah there is probably some trick, but first things first I would like to know if you could sub for constants or not, because otherwise Im just going down the wrong road

Answer 18

no i don't think so, i think you sub for the variable

Answer 19

let me try with pencil and paper and see if something nice comes out of it

Answer 20

what is alpha?

Answer 21

The first question i did like this
a=tan(alpha), x=tan(beta)
alpha=tan^-1(a), beta=tan^-1(x)
y=tan^-1(tan(alpha)+tan(beta)/1-tan(alpha)tan(beta))
y=tan^-1(tan(alpha+beta))
y=alpha+beta
y=tan^-1(a)+tan^-1(x)
dy/dx=1/1+x^2

Answer 22

alpha is variable of substitution I guess, doesn't really matter as we would sub back alpha for x

Answer 23

i think you best leave \(a\) alone as it is a constant, but i could be wrong

Answer 24

hmm i think that works
...
http://www.wolframalpha.com/input/?i=+differentiate+arctan%28%28a%2Bx%29%2F%281-ax%29%29

yep:)

Answer 25

Yeah I was confused about that, but you can see how easily a solution comes out if you can sub for constants...

Answer 26

your answer is definitely right

Answer 27

yeah i guess you can
the constants drop out

Answer 28

Yeah, I checked the answer. It IS right but I don't know if the method IS right

Answer 29

@misty1212 , rewriting a constant in different form is always ok
k = e^C kinda thing

Answer 30

so trick was recognizing the addition angel formula for tangent

Answer 31

yeah, it looks like the tan(2x) identity if you notice carefully

Answer 32

nice job, i didn't notice that at all

Answer 33

OH! light goes on
learn something new every day

Answer 34

sorry not tan(2x), tan(x+y)

Answer 35

yeah got that
wow

Answer 36

so for the second one, use your answer to first one plus chain rule right? or is there some other thing going on?

Answer 37

you probably factor out a or b from both numberator and denominator then sub

Answer 38

something like 1+(b/a)tanx, something like that

Answer 39

sorry i was away, yes make that sub ---> b/a = tan(alpha)

use same identity as last one
then use chain rule when doing derivative

answer is 1

Answer 40

@misty1212 @dumbcow
Yeah I solved it, took out b from both numerator and denominator
\[y=\tan^-1(\frac{ b(\frac{ a }{ b }+tanx }{ b(1-\frac{ atanx }{ b }) }) \]
after that \[ \frac{ a }{ b }=\tan \alpha\]
\[\alpha=\tan^-1\frac{ a }{ b }\]
\[y=\tan^-1(\frac{ \tan \alpha+tanx }{ 1-\tan \alpha tanx })\]
\[y=\tan^-1(\tan (\alpha+x))\]
\[y=\alpha+x\]\[y=\tan^-1\frac{ a }{ b }+x\]\[\frac{ dy }{ dx }=1\]

Answer 41

:)