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

1 Answer
Somebody N.
Mar 17, 2018

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

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