1. Another Differentiation Problem

Find dy/dx for the following curve:

x = ( 1 - a sq ) to the power of 3/2
y = sin to the power of -1 ( a )

2. can you confirm the equations

Looks like,

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

3. Originally Posted by puggie
Find dy/dx for the following curve:

x = ( 1 - a sq ) to the power of 3/2
y = sin to the power of -1 ( a )
As I said in your last post, this requires Parametric Differentiation.

Remember, $\frac{dy}{dA} \times \frac{dA}{dx} = \frac{dy}{dx}$

Is it differentiating the equations themselves you are struggling with?

4. Hi pickslides

y as you have written is correct.

x should be a square and not square root of a!

Thanks everybody

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

Substitute $a=sin \theta$

$y=\theta$

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

$x=cos^3\theta$

$\frac{dx}{d\theta}=3cos^2\theta*-sin\theta$
$\frac{dx}{dy}=3cos^2\theta*-sin\theta$
$\frac{dx}{dy}=3(1-a^2)*-a$
$\frac{dx}{dy}=3a(a^2-1)$
$\frac{dy}{dx}=\frac{1}{3a(a^2-1)}$

6. $x= (1-a^2)^{\frac{3}{2}}$

from this find $\frac{dx}{da}$

$y= sin^{-1}(a)$

from this find $\frac{dy}{da}$

$then \frac{dy}{dx} = \frac{dy}{da} \times \frac{1}{\frac{dx}{da}}
$