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,

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

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

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

Substitute $\displaystyle a=sin \theta$

$\displaystyle y=\theta$

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

$\displaystyle x=cos^3\theta$

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

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

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

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

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

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