Question

what is the derivative of (x^2+1)^(arcsin(x))?

Answer 1

having trouble on what to do.

Answer 2

what do you mean by taking the log

Answer 3

for the second method i got e^((x2+1)ln(arcsin(x))) and then the derivative of (x2+1)ln(arcsin(x)

Answer 4

i dont understand how to do the first method however

Answer 5

wait how do you take the derivative of a log though

Answer 6

it is a product rule and also a chain rule problem

Answer 7

ehhh you're talking about the second method right?

Answer 8

the derivative of
\(\ln(f(x))\) is \(\frac{f'(x)}{f(x)}\)

either method, it amounts to exactly the same work

Answer 9

when you're saying you take the log of both sides you mean ln(y)=ln(x^2+1)^(arcsinx) right and not log(y)=log(...)

Answer 10

i know how to take the derivative of ln(x) its 1/x times x'

Answer 11

it is
\[\ln(y)=\] oh damn i had it backwards!!

Answer 12

\[y=(x^2+1)^{\arcsin(x)}\]
\[\ln(y)=\arcsin(x)\ln(x^2+1)\] thats better, sorry

Answer 13

now take the derivative using the chain rule and product rule

Answer 14

ok so i get 1/y = dy/dx arc sin(x) (1/x^2+1)(2x)

Answer 15

and i divide everything by arc sin(x)(1/x^2+1)(2x) to have dy/dx on one side and i get arc sin(x)(1/x^2+1)(2x)/y = dy/dx?

Answer 16

\[\ln(y)=\arcsin(x)\ln(x^2+1)\]
\[\frac{y'}{y}=\frac{1}{\sqrt{1-x^2}}\ln(x^2+1)+\arcsin(x)\frac{2x}{x^2+1}\]

Answer 17

product rule and chain rule

Answer 18

ok thank you

Answer 19

yw
then to finish multiply all this mess by the original function