Question

Can someone explain how to differentiate this? i know the answer i just don't understand the process.
Find the derivative of the function, f(x)=tan^-1sqrt(3x+1)

Answer 1

so is st just this particular rpoblem or inverse trig functions in general?

Answer 2

inverse trig functions in general

Answer 3

so let's say you have \(y=\sin^{-1}x\). first let's rewrite so we have something we know how to differentiate...\[y=\sin^{-1}x\Rightarrow \sin y = x\]
now, differntiate implicitly

Answer 4

ooh i see, 1/cos is sec. but why does it equal \[\sec(\sin ^{-1}x)\]

Answer 5

\[(\cos y)\frac{ dy }{ dx }=1\Rightarrow\frac{ dy }{ dx }=\frac{ 1 }{ \cos y }\]

Answer 6

sorry, you don't want to go that way. (but \(\frac{1}{\cos x}=\sec x\)
anyway, we know \(\sin^2 x+ \cos^2 x = 1\Rightarrow \cos x = \sqrt{1-\sin^2 x}\)

Answer 7

ok i understand that

Answer 8

so we get \[\frac{ dy }{ dx }=\frac{ 1 }{ \cos y }=\frac{ 1 }{ \sqrt{1-\sin^2 y} }=\frac{ 1 }{ \sqrt{1-\sin^2 (\sin^{-1}x)} }=\frac{ 1 }{ \sqrt{1-x^2} }\]

Answer 9

remember, \(\sin^2 x = (\sin x)(\sin x) \text{ and } \sin(\sin^{-1} x)=\sin^{-1}(\sin x) = x\)

Answer 10

i'm trying but i don't understand where \[\sin ^{-1}x\] comes from

Answer 11

that's y. remember at the beginning that \(y=\sin^{-1}x\)?

Answer 12

ooh okay got it

Answer 13

you do the same for all of the inverse trig functions and then use one of the pythagorean identities to get back x, like we did here.

Answer 14

okay that makes sense, But how does that translate into my original question when there is no y?

Answer 15

http://www.math.brown.edu/utra/trigderivs.html

here is a site that should also help explain the derivatives of inverse trig functions.

now, in your question, you have \[f(x)=\tan^{-1}\sqrt{3x+1}\]f(x) is y. anyways, from deriving it yourself or a table of derivatives, \[\frac{ d }{ dx }\tan^{-1}x = \frac{ 1 }{ 1+x^2 }\]

you will have to use the chain rule because of the function inside.

Answer 16

remember, y = f(x)

Answer 17

yeah i got that i understand up until after \[\frac{ 1 }{ \sqrt{3x+1} } \times {3x+1}\]

Answer 18

i also use chegg and i have this but i dont understand how they got to the second step and beyond

Answer 19

\[\frac{ d }{ dx }\tan^{-1}\left( \sqrt{3x+1} \right)=f'\left( x \right)=\frac{ 1 }{1+\left( \sqrt{3x+1} \right)^2 }\cdot \frac{ 1 }{ 2\sqrt{3x+1} }\cdot 3\]

Answer 20

so\[f'\left( x \right)=\frac{ 1 }{ 1+3x+1 }\cdot \frac{ 1 }{ 2\sqrt{3x+1} }\cdot 3=\frac{ 3 }{ 2\left( 3x+2 \right)\sqrt{3x+1} }\]

Answer 21

ooh alright, I understand it now. Thank you.

Answer 22

you're welcome

Answer 23

if you're good, you should close the question