Question

calculus help! find an equation of the tangent to the graph of y=arcsin(5x) at the point (1/5(Sq. root of 2), pi/4)

Answer 1

ick

Answer 2

yeah

Answer 3

Satellite, please, help here: http://openstudy.com/study#/updates/507cd0dfe4b07c5f7c1fb4c3

Answer 4

the answer is supposed to be y=5(sq rt 2x)+pi/4-1

Answer 5

derivative is easy enough, it is
\[\frac{5}{\sqrt{1-25x^2}}\]

Answer 6

ok

Answer 7

replace \(x\) by \(\frac{1}{5\sqrt{2}}\) and see what you get for the slope

Answer 8

can u explain this to be im kinda lost :l

Answer 9

how to get the derivative, or why to evaluate ?

Answer 10

how to get the derivative

Answer 11

:l

Answer 12

1) memorize the derivative of arcsine
it is \(\frac{1}{\sqrt{1-x^2}}\)
2) use the chain rule

Answer 13

i plug in 5x for x right?

Answer 14

so since you have
\[\arcsin(5x)\] you replace \(x\) by \(5x\) and then multiply by the derivative of \(5x\) which is just 5
that gives you
\[\frac{1}{\sqrt{1-(5x)^2}}\times 5=\frac{5}{\sqrt{1-25x^2}}\]

Answer 15

crystal clear! :}

Answer 16

actually first expression is just as easy to evaluate as the second so you can leave it like that
rest is more or less routine right?
find the slope by evaluating the derivative at the first coordinate
use the point slope formula
finito

Answer 17

beautiful! thanks so much!

Answer 18

wait but im not getting what im suppose to get :[