Differentiating harder inverse trig (1 Viewer)

[Danmiff]

hey
just wondering how you'd go about differentiating:
y = sin^-1 f(x)

where f(x) is a function with power of x that DOES NOT = 1

for example... y = sine^-1 [(x)^(1/2)]

cheers

 

[khorne]

You can use the chain rule, and in this case, make a substitution u = rt(x):

so:

d/dx arcsin(x^.5) = d arcsin(u)/du *du/dx (d/dx arcsin(u)/du = 1/rt[1-u] )

=d/dx rt(x)/rt(1-x) = 1/2rt(1-x) rt(x)

 

[lychnobity]

EDIT:

 

[scardizzle]

it' s kind of like the chain rule:

d (sin^-1 f(x)) = 1/(1- f(x)^2)^1/2 x f'(x)
dx

 

[tom.evans.15]

it' s kind of like the chain rule

Correct me if im wrong, but it IS the chain rule right?

y=asin(f(x)) is in the form y=g(f(x)) where g(x) = asinx

therefore dy/dx = g'(f(x)) * f'(x)

so d/dx(asinf(x)) = (1/sqrt(1-(f(x)^2))*(f'(x))

and please, correct me if im wrong

 

[khorne]

Yes, you're right, but 3 people beat you to it =]

 

[tom.evans.15]

Yes, you're right, but 3 people beat you to it =]

Oh well. Doesn't matter...

tbh, the explanation was more for my own sake rather than OPs or anyone elses...

but i'm right, so i feel good now

 

[scardizzle]

well i only said it was kinda of like the chain rule because my idea of the chain rule is

d(f(x)^n) = n f(x)^n-1 . f'(x)
dx

so i guess maybe i should have said this was a different version of the chain rule

 

[untouchablecuz]

imo



is the easiest/most useful form of the chain rule