Question

find the derivative of the following function:
y=37 arctan (sqrtx)..
Please show step by step

Answer 1

pull the constant; and chain the rest

Answer 2

as if....

Answer 3

can you show me

Answer 4

RMALIK HAVENT U UNDERSTOOD CHAIN RULE YET? i mean weve done arnd half a dozen qstns for u

Answer 5

alright

Answer 6

no need to berate ....

Answer 7

so do i explain again?

Answer 8

we explain as many times as it takes :) imo

Answer 9

and again and again until u finish all the calculus for one lifetime?

Answer 10

well isnt what this site for? to help

Answer 11

yes, but its hard at times to tell who is abusing the site; and who is using it

Answer 12

im not abusing the site i just like to look at steps of the problem

Answer 13

fyn

1/2(1+x)sqrt (x)

Answer 14

u treat rootx as x first
and then write the derivative

Answer 15

1/(1 + rootx ^2)

Answer 16

now multiply it by the derivative of rootx

Answer 17

tan-1(x) = y means;

x = tan(y) ; derive both sides to get

1 = sec^2(y) y'

y' = 1/sec^2(y) ; but y = tan-1(x) soo...

y' = 1\(sec^2(tan-1(x))) which can be simplified right?

Answer 18

yes where tan becomes sin over cos

Answer 19

Personally I like to use a substitution when explaining the chain rule. Since it's often shown as:
\[f(g(x))' = f'(x)g'(x)\]

So we just need to pick our f and g.

Answer 20

its been awhile since I tried to step thru it :)

Do we translate it from the standard triangle format for it?

Answer 21

In this case f(a) = 37arctan(a) and g(x) = sqrt(x)

Answer 22

ok polpak thats good explaination

Answer 23

I'm sure you can probably find \(g'(x)\) pretty easily by now. So the only question is what's the derivative of arctan(a) ?

Answer 24

tan(y) = x/1

sec(y) = sqrt(x^2 +1)

sec^2(y) = x^2 +1

Dx(tan-1(x)) = 1/(x^2+1) .... right?

Answer 25

My guess is you have that in a table you're supposed to memorize, but yes amistre64 that's correct.

Answer 26

yay!! ... that was easier to remember than I thought :) thnx

Answer 27

Then you just multiply the two derivatives (and simplify if possible)