Question

How do you evaluate the partial derivative of the following sine expression?

`sin(2\alpha)=\frac{2\sigma_{z\theta}}{\sigma_1-\sigma_3}`

`\frac{\partial\alpha}{\partial\sigma_1}=?`

`\frac{\partial\alpha}{\partial\sigma_3}=?`

Answer 1

Given:

`sin(2alpha)=(2sigma_(ztheta))/(sigma_1-sigma_3)`

Solve for `alpha`

`2alpha=sin^-1((2sigma_(ztheta))/(sigma_1-sigma_3))`

`alpha=1/2sin^-1((2sigma_(ztheta))/(sigma_1-sigma_3))`

Both derivatives are merely the chain rule:

Let `u = (2sigma_(ztheta))/(sigma_1-sigma_3)`

`(d(1/2sin^-1(u)))/(du) = 1/2 1/sqrt(1-u^2)`

`(delu)/(delsigma_1) = -(2sigma_(ztheta))/(sigma_1-sigma_3)^2`

`(delu)/(delsigma_3) = (2sigma_(ztheta))/(sigma_1-sigma_3)^2`

Putting the chain rule together with respect to `sigma_1`:

`(del alpha)/(del sigma_1) = 1/2 1/sqrt(1-((2sigma_(ztheta))/(sigma_1-sigma_3))^2)(-2sigma_(ztheta))/((sigma_1-sigma_3)^2)`

Simplify:

`(del alpha)/(del sigma_1) = -sigma_(ztheta)/((sigma_1-sigma_3)sqrt((sigma_1-sigma_3)^2-(2sigma_(ztheta))^2))`

Putting the chain rule together with respect to `sigma_3`:

`(del alpha)/(del sigma_3) = 1/2 1/sqrt(1-((2sigma_(ztheta))/(sigma_1-sigma_3))^2)(2sigma_(ztheta))/((sigma_1-sigma_3)^2)`

Simplify:

`(del alpha)/(del sigma_3) = sigma_(ztheta)/((sigma_1-sigma_3)sqrt((sigma_1-sigma_3)^2-(2sigma_(ztheta))^2))`