How do you differentiate #(x-1)/sqrt(6-x)#?

1 Answer
Jan 23, 2018

By using quotient rule:
enter image source here

Explanation:

Let #u=x-1# and #v=sqrt(6-x)#
#(du)/dx = 1# and #(dv)/dx = -1/2(6-x)^(-1/2)#

Plugging this into the formula, you get:

Step one:

#dy/dx = ((sqrt(6-x))(1)-(x-1)(-1/2)(6-x)^(-1/2))/((sqrt(6-x))^2)#

Step two: simplify

#dy/dx = ((sqrt(6-x))+(1/2)(x-1)(6-x)^(-1/2))/(6-x)#

Step three: split numerator

#dy/dx = (sqrt(6-x))/(6-x) + ((1/2)(x-1)(6-x)^(-1/2))/(6-x)#

Step Four: combine the #(6-x)# terms using exponent rules

#dy/dx = (6-x)^(-1/2) + (1/2)(x-1)(6-x)^(-3/2)#

Step Five: write with positive exponents only

#dy/dx = 1/((6-x)^(1/2)) + (x-1)/(2(6-x)^(3/2))#

There may be other ways to further simplify the derivative, but this is what I would have done.

For future questions like this, use https://www.derivative-calculator.net/ It gives you the step by step process for any differentiation.