Question

How do I derive `y=sqrt(9+x^2)+sqrt(x^2-10x+41)`?

Answer 1

`color(blue)(=(x)/(sqrt(9+x^2))+(x-5)/sqrt(x^2-10x+41))`

Explanation:

We need to use the chain rule here:

`dy/dx=dy/(du)*(du)/dx`

In order not to use two variables when substituting, we can differentiate in parts:

`dy/dx` is distributive over the sum, so:

`dy/dx(sqrt(9+x^2)+sqrt(x^2-10x+41))`

`=dy/dx(sqrt(9+x^2))+dy/dx(sqrt(x^2-10x+41))`

For the first term:

Rewriting

`sqrt(9+x^2)=(9+x^2)^(1/2)`

let: `u=9+x^2`

`:.`

`y=u^(1/2)`

Using the chain rule:

`dy/dx=1/2u^(-1/2)*(2x)=(2x)/(2sqrt(u))=(x)/(sqrt(9+x^2))`

For the second term:

`dy/dx=dy/(dv)*(dv)/dx`

Rewriting `sqrt(x^2-10x+41)=(x^2-10x+41)^(1/2)`

Let: `v=x^2-10x+41`

Using the chain rule:

`dy/dx=1/2v^(-1/2)*(2x-10)=(2x-10)/(2sqrt(v))=(x-5)/sqrt(x^2-10x+41)`

Combining the two parts:

`dy/dx=color(blue)((x)/(sqrt(9+x^2))+(x-5)/sqrt(x^2-10x+41))`

I would leave this in this form. If you add the fractions it will make for a messy expression