Finding the Derivative of a Square Root Function

[JackieAnne]

Hi so I am new to calculus and need help with differentiation

The question is:
Differentiate the following function:

sqrt(x^2-8x+20)

So far I have gotten
=(x^2-8x+20)^(1/2)
=(1/2)(x^2-8x+20)^(-1/2) * 2x-8

I am unsure where to go from here
I know we are supposed to use the chain rule but I am still confused where to go after this step.

Thanks

 

[Mark44]

Hi so I am new to calculus and need help with differentiation

The question is:
Differentiate the following function:

sqrt(x^2-8x+20)

So far I have gotten
=(x^2-8x+20)^(1/2)
=(1/2)(x^2-8x+20)^(-1/2) * 2x-8

I am unsure where to go from here
I know we are supposed to use the chain rule but I am still confused where to go after this step.

Thanks

You have
sqrt(x^2-8x+20)
=(x^2-8x+20)^(1/2)
Up to here, what you have done is rewrite the expression in a different form.

What you have below is the derivative of the expression above. It is not equal to what you started with, so you should not have =. This is the derivative, and it is mostly correct (you need parentheses around 2x-8.

=(1/2)(x^2-8x+20)^(-1/2) * 2x-8

Be sure to distinguish between the function you're differentiating and what you get when you differentiate (the deriviative). Here's one way:

y = sqrt(x^2-8x+20) = (x^2-8x+20)^(1/2)

dy/dx = (1/2)(x^2-8x+20)^(-1/2) * (2x-8)
= (1/2)(2x - 8)/(x^2 - 8x + 20)^(1/2) = (x - 4)/(x^2 - 8x + 20)^(1/2)

You can also write dy/dx as y'.

 

[JackieAnne]

Thank-you!