Question on differentiation of implicit functions (1 Viewer)

Skeptyks · Oct 9, 2011

EDIT: Woops, the topic was differentiation of implicit functions but I just realised that this one isn't one of them... My bad, didn't mean for the misleading title.

Sorry for posting these easy questions but I can't figure out this one:
Given that $y=\sin^{-1}\sqrt{x}$ prove that $\frac{dy}{dx}=\frac{1}{sin2y}$ .

nightweaver066 · Oct 9, 2011

$siny = \sqrt{x}$
$\text{Differentiating both sides with respect to x}$
$\frac{dy}{dx}cosy = \frac{1}{2\sqrt{x}}$
$\frac{dy}{dx} = \frac{1}{2cosy\sqrt{x}}$
$\text{Now remembering that } siny = \sqrt{x}$
$\frac{dy}{dx} = \frac{1}{2cosysiny}$
$\frac{dy}{dx} = \frac{1}{sin2y}$

AAEldar · Oct 9, 2011

As above ^.

Ninja'd >.>

EDIT: Actually I did it differently. Differentiate the inverse function and times it by the derivative of the inside and sub in for x. Little algebra and it comes out the same.

D94 · Oct 9, 2011

You may have forgotten to multiple by the derivative of $\sqrt{x}$ .

Skeptyks · Oct 9, 2011

nightweaver066 said:
$siny = \sqrt{x}$
$\text{Differentiating both sides with respect to x}$
$\frac{dy}{dx}cosy = \frac{1}{2\sqrt{x}}$
$\frac{dy}{dx} = \frac{1}{2cosy\sqrt{x}}$
$\text{Now remembering that } siny = \sqrt{x}$
$\frac{dy}{dx} = \frac{1}{2cosysiny}$
$\frac{dy}{dx} = \frac{1}{sin2y}$

Thanks :] I hate it when there are no worked solutions for these questions.

Question on differentiation of implicit functions (1 Viewer)

Skeptyks

Member

nightweaver066

Well-Known Member

AAEldar

Premium Member

D94

New Member

Skeptyks

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)