derivative of inverse trig function

[fais_attention]

f(x) = 8 sin[sup:149hgwo7]-1[/sup:149hgwo7](x[sup:149hgwo7]3[/sup:149hgwo7])

find f ' (x).

First I took the exponent 3 and moved it down to be a coefficient. That might be my first mistake. So, I have 24arcsin(x)=f(x) I figured, hey, that must be
24/(sqrt(1-x^2)). But, that is wrong.

If someone would please point out my mistake, I can probably take it from there.
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Edited by stapel -- Reason for edit: Replacing error-prone hotlink to image of text on secure (?) server with actual text.

 

[stapel]

Given f(x) = 8 sin[sup:3o68xvel]-1[/sup:3o68xvel](x[sup:3o68xvel]3[/sup:3o68xvel]), find f ' (x).

First I took the exponent 3 and moved it down to be a coefficient.

You might want to review the Chain Rule. You're supposed to work from the outside in, not the inside out. :wink:

Eliz.

 

[galactus]

Try a sub like in integration.

Let \(\displaystyle u=x^{3}, \;\ du=3x^{2}dx\)

Make the subs and you may see it better.

\(\displaystyle \frac{d}{du}[8sin^{-1}(u)]\)

Then, you get \(\displaystyle \frac{8}{\sqrt{1-u^{2}}}du\)

Now, can you see it?.

 

[fais_attention]

Thanks! I figured it out, and it was 24x^2 / sqrt(1-x^6).

 

[Deleted member 4993]

f(x) = 8 sin[sup:24bxb0b9]-1[/sup:24bxb0b9](x[sup:24bxb0b9]3[/sup:24bxb0b9])

find f ' (x).

First I took the exponent 3 and moved it down to be a coefficient. That might be my first mistake. So, I have 24arcsin(x)=f(x) I figured, hey, that must be
24/(sqrt(1-x^2)). But, that is wrong.

If someone would please point out my mistake, I can probably take it from there.
___________________________
Edited by stapel -- Reason for edit: Replacing error-prone hotlink to image of text on secure (?) server with actual text.

\(\displaystyle y \, = \, 8\sin^{-1}(x^3)\)

\(\displaystyle x^3 \, = \, sin(\frac{y}{8})\)................\(\displaystyle \sqrt{1-x^6} \, = \, cos(\frac{y}{8})\)

differentiate w.r.t. x

\(\displaystyle 3\cdot x^2 = \, \frac{1}{8}\cdot cos(\frac{y}{8})\frac{dy}{dx}\)

\(\displaystyle \frac{dy}{dx} \, = \, \frac{24\cdot x^2}{\sqrt{1-x^6}}\)