Question

Derivative of sin inverse 2^x+1/1+4^x

Answer 1

\[\sin^{-1} \left( 2^x+1\div4^x+1\right)\]

Answer 2

please help :'(

Answer 3

Alright, well do you know the derivative of just y=sin^-1(x) ?

After that, you're just applying the chain rule.

Answer 4

can u please solve it... i cnt do it :(

Answer 5

?

Answer 6

ny1 ?

Answer 7

I mean if you can't do it, you're not gonna have a fun time on the test. I won't be taking your test for you.

Answer 8

dude...plz...its kinda urgent :'(

Answer 9

Kainui already asked this but you didn't answer for some reason.
Do you know the derivative of \(\large sin^{-1} x\) ?

Answer 10

yes... 1/under root1+x^2

Answer 11

i tried it lyk 10 times... please help :/ i did apply the chain rule..still m not gettin the solution

Answer 12

\[\large \frac{d}{dx}\sin^{-1}\color{orangered}{x} \qquad=\qquad \frac{1}{\sqrt{1-\color{orangered}{x}^2}}\]

I think it's subtraction in the bottom, but ok that's a good start :)
Let's apply that to our problem.

Answer 13

\[\large \frac{d}{dx}\sin^{-1}\color{orangered}{\left(\frac{2^x+1}{4^x+1}\right)} \qquad=\qquad \frac{1}{\sqrt{1-\color{orangered}{\left(\dfrac{2^x+1}{4^x+1}\right)}^2}}\color{royalblue}{\left(\frac{2^x+1}{4^x+1}\right)'}\]

Follow along so far?
The blue term shows up due to the chain rule.
The prime is to show us that we still need to take the derivative of that part.

Answer 14

yes i got it.. now what?

Answer 15

Now we have to apply the `Quotient Rule` to the blue part.

Answer 16

Here is the setup for the quotient rule,
\[\large \left(\frac{2^x+1}{4^x+1}\right)'=\frac{\color{royalblue}{(2^x+1)'}(4^x+1)-(2^x+1)\color{royalblue}{(4^x+1)'}}{(4^x+1)^2}\]

The blue terms are the ones we have to take derivatives of.
Understand how that was setup?

Answer 17

yes..now ?

Answer 18

Do you remember your exponential derivatives? They're a lil tricky.

What's the derivative of \(\large 2^x+1\) ?

Answer 19

2^x / log2

Answer 20

mmm not division :) just multiplication, but yes good.

\[\large (2^x)'=2^x(\ln2)\]

Answer 21

now?

Answer 22

\[\large \frac{\color{royalblue}{(2^x+1)'}(4^x+1)-(2^x+1)\color{royalblue}{(4^x+1)'}}{(4^x+1)^2}\]

So taking the derivative of that first blue term gives us,\[\large \frac{\color{orangered}{(2^x(\ln2)+0)}(4^x+1)-(2^x+1)\color{royalblue}{(4^x+1)'}}{(4^x+1)^2}\]

What do you get for the other blue part?

Answer 23

4^x log 4

Answer 24

\[\large \frac{\color{orangered}{2^x(\ln2)}(4^x+1)-(2^x+1)\color{orangered}{4^x(\ln4)}}{(4^x+1)^2}\]

Ok good, let's put this together with our first part that we did.

\[\frac{d}{dx}\sin^{-1}\color{orangered}{\left(\frac{2^x+1}{4^x+1}\right)} =\frac{1}{\sqrt{1-\color{orangered}{\left(\dfrac{2^x+1}{4^x+1}\right)}^2}}\color{royalblue}{\left(\frac{2^x+1}{4^x+1}\right)'}\]

So we did the blue part, giving us,\[\frac{d}{dx}\sin^{-1}\color{orangered}{\left(\frac{2^x+1}{4^x+1}\right)} =\frac{1}{\sqrt{1-\color{orangered}{\left(\dfrac{2^x+1}{4^x+1}\right)}^2}}\color{royalblue}{\left(\large \frac{2^x(\ln2)(4^x+1)-(2^x+1)4^x(\ln4)}{(4^x+1)^2}\right)}\]

Answer 25

ah it got cut off.. sec ill paste it again,

Answer 26

ya sure

Answer 27

\[\large \frac{1}{\sqrt{1-\color{orangered}{\left(\dfrac{2^x+1}{4^x+1}\right)}^2}}\color{royalblue}{\left(\large \frac{2^x(\ln2)(4^x+1)-(2^x+1)4^x(\ln4)}{(4^x+1)^2}\right)}\]

Answer 28

I mean.. I guess you could simplify ... maybe..
But I don't see the point.

That's your answer right there.

Answer 29

thankyou :)

Answer 30

np c: