Inverse Trig. Differentiation (1 Viewer)

SunnyScience · Apr 11, 2012

Can someone please show me how to find the derivative for equations such as this please!

y = sin^-1 [(2x-1)/3]

Thanks!

nightweaver066 · Apr 11, 2012

$y = sin^{-1}(\frac{2x - 1}{3})$

$$Using: $ \frac{d}{dx}[sin^{-1}(f(x))] = \frac{1}{\sqrt{1 - [f(x)]^2}} \times f'(x)$

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (\frac{2x - 1}{3})^2}} \times \frac{2}{3}$

$= \frac{3}{\sqrt{9 - 4x^2 + 4x - 1}} \times \frac{2}{3}$

$= \frac{3}{\sqrt{8 + 4x - 4x^2}} \times \frac{2}{3}$

$= \frac{2}{2\sqrt{2 + x - x^2}}$

$= \frac{1}{\sqrt{2 + x - x^2}}$

SunnyScience · Apr 11, 2012

nightweaver066 said:
$y = sin^{-1}(\frac{2x - 1}{3})$

$$Using: $ \frac{d}{dx}[sin^{-1}(f(x))] = \frac{1}{\sqrt{1 - [f(x)]^2}} \times f'(x)$

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (\frac{2x - 1}{3})^2}} \times \frac{2}{3}$

$= \frac{3}{\sqrt{9 - 4x^2 + 4x - 1}} \times \frac{2}{3}$

$= \frac{3}{\sqrt{8 + 4x - 4x^2}} \times \frac{2}{3}$

$= \frac{2}{2\sqrt{2 + x - x^2}}$

$= \frac{1}{\sqrt{2 + x - x^2}}$

Thanks!

One quick question (though it is absolutely retarded and my maths teacher would shoot me for asking) - how do you get from:

$= \frac{1}{\sqrt{1 - (\frac{2x - 1}{3})^2}} \times \frac{2}{3}$

to

$= \frac{3}{\sqrt{9 - 4x^2 + 4x - 1}} \times \frac{2}{3}$

? :/

nightweaver066 · Apr 11, 2012

SunnyScience said:
Thanks!

One quick question (though it is absolutely retarded and my maths teacher would shoot me for asking) - how do you get from:

$= \frac{1}{\sqrt{1 - (\frac{2x - 1}{3})^2}} \times \frac{2}{3}$

to

$= \frac{3}{\sqrt{9 - 4x^2 + 4x - 1}} \times \frac{2}{3}$

? :/

Sorry that i skipped steps lol.

What i did was expand the [(2x - 1)/3]^2, combine the two terms on the bottom to have a common denominator, took out 1/root(9) from the bottom, moved it to the top and changed it to 3.

SunnyScience · Apr 11, 2012

Also, could you use:

d/dx (sin^-1 x/a) = 1/[(a^2 - x^2)^1/2]

to solve this question?

nightweaver066 · Apr 11, 2012

SunnyScience said:
Also, could you use:

d/dx (sin^-1 x/a) = 1/[(a^2 - x^2)^1/2]

to solve this question?

Yep you can.

SunnyScience · Apr 11, 2012

Thank you for your time

SunnyScience · Apr 12, 2012

Sorry, but can anyone please show me how to solve these equations using this rule please:

d/dx (sin^-1 x/a) = 1/[(a^2 - x^2)^1/2]

I'm just not getting that ^^^^

pokka · Apr 12, 2012

SunnyScience said:
Sorry, but can anyone please show me how to solve these equations using this rule please:

d/dx (sin^-1 x/a) = 1/[(a^2 - x^2)^1/2]

I'm just not getting that ^^^^

$$We want to manipulate our expression so it is in the form $y=\sin^{-1} \left( \frac{x-b}{a}\right )$ if we want to use that formula (the formula still works when you replace $x$ with $x-b$ as long it is a monic linear function):$\\\\\indent y = \sin^{-1} \left( \frac{2x-1}{3}\right )\\\\\indent\:\:=\sin^{-1} \left( \frac{2(x-\frac{1}{2})}{3}\right )\\\\\indent\:\:=\sin^{-1} \left( \frac{2(x-\frac{1}{2})}{3}\right) \\\\\indent y=\sin^{-1} \left( \frac{x-\frac{1}{2}}{\frac{3}{2}}\right)\\\\$Now we can differentiate using $\frac{d(\sin^{-1}\frac{(x-b)}{a})}{dx}=\frac{1}{\sqrt{a^2-(x-b)^2}}{}:\\\\\frac{dy}{dx}=\frac{1}{\sqrt{\frac{9}{4}-(x-\frac{1}{2})^2}}\\\\\indent =\frac{1}{\sqrt{\frac{9}{4}-(\frac{4x^2-4x+1}{4})}}\\\\\indent =\frac{2}{\sqrt{8+4x-4x^2}}\\\\\indent = \frac{2}{2\sqrt{2+x-x^2}}\\\\\indent =\frac{1}{\sqrt{2+x-x^2}}$

SunnyScience · Apr 12, 2012

pokka said:
$$We want to manipulate our expression so it is in the form $y=\sin^{-1} \left( \frac{x-b}{a}\right )$ if we want to use that formula (the formula still works when you replace $x$ with $x-b$ as long it is a monic linear function):$\\\\\indent y = \sin^{-1} \left( \frac{2x-1}{3}\right )\\\\\indent\:\:=\sin^{-1} \left( \frac{2(x-\frac{1}{2})}{3}\right )\\\\\indent\:\:=\sin^{-1} \left( \frac{2(x-\frac{1}{2})}{3}\right) \\\\\indent y=\sin^{-1} \left( \frac{x-\frac{1}{2}}{\frac{3}{2}}\right)\\\\$Now we can differentiate using $\frac{d(\sin^{-1}\frac{(x-b)}{a})}{dx}=\frac{1}{\sqrt{a^2-(x-b)^2}}{}:\\\\\frac{dy}{dx}=\frac{1}{\sqrt{\frac{9}{4}-(x-\frac{1}{2})^2}}\\\\\indent =\frac{1}{\sqrt{\frac{9}{4}-(\frac{4x^2-4x+1}{4})}}\\\\\indent =\frac{2}{\sqrt{8+4x-4x^2}}\\\\\indent = \frac{2}{2\sqrt{2+x-x^2}}\\\\\indent =\frac{1}{\sqrt{2+x-x^2}}$

Thank you sooooo much - it was really bugging me

Just a quick question, what would you do if the numerator had perhaps a non-linear equation, such as a quadratic or a cubic? And like with the denominator?

pokka · Apr 12, 2012

then you would use the chain rule like nightweaver066 did

Inverse Trig. Differentiation (1 Viewer)

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