How do you find the derivative of #xsqrt (1-x)#?

Question

Samantha Adkison · Accepted Answer

#(2-3x)/(2sqrt(1-x)# #color(red)(d/(dx)(u*v)=u*(dv)/(dx)+v*(du)/(dx))# and,#d/(dx)(sqrt(1-x))=d/(dx)(1-x)^(1/2)=1/2*(1-x)^(-1/2)=1/2*1/sqrt(1-x)# #y=x*sqrt(1-x)# #=>(dy)/(dx)=x*d/(dx)(sqrt(1-x))+sqrt(1-x)*d/(dx)(x)# #=>(dy)/(dx)=x*1/(2sqrt(1-x))(-1)+sqrt(1-x)*1# #=(-x)/(2sqrt(1-x))+sqrt(1-x)=(-x+2(1-x))/(2sqrt(1-x)# #=(-x+2-2x)/(2sqrt(1-x))=(2-3x)/(2sqrt(1-x)#

Christopher Agresta · Answer

#(2- 3x)/(2 sqrt(1-x))#

The expression is product of two functions in #x#.

Denoting these by #f(x)# and #g(x)#, respectively,

the first is #f(x) = x#

and the second is #g(x) = sqrt(1 - x)#

#g(x)# is a compound function (ie ie a function of a function)

The derivative of the expression is

#f'(x)g(x) + f(x)g'(x)#

The derivative of the first function is straightforward #f'(x) = 1#

The derivative of the second is trickier because it is a compound function. This requires the chain rule. The outer function is the square root function, and the inner function is the polynomial #(1 - x)#

writing the compound function as #h(j(x))# (#h# of #j# of #x#), the derivative is

#h'(j(x))j'(x)#

That is, the derivative of the outer function evaluated at the inner function times the derivative of the inner function

It makes things simpler to rewrite #g(x)# using index notation, that is

#g(x) = (1 - x)^(1/2)#

Evaluating the outer function is the straightforward application of the rules of polynomial differentiation applied to its index, that is

#h'(j(x)) = 1/2(1-x)^(1/2 - 1) = 1/2(1-x)^(-1/2)#

And the derivative of the inner function is #j'(x) = -1#

So the derivative of the compound function #g(x)# is

#g'(x) = h'(j(x))j'(x) = 1/2(1-x)^(-1/2)(-1)# #= -1/2 (1 - x )^(-1/2)#

Or, if you prefer, reverting to the square root notation and noting the negative index

#g'(x) = - 1/(2 sqrt(1-x))#

So the overall derivative is

#f'(x)g(x) + f(x)g'(x)#

#= (1)(1 - x)^(1/2)+ (x)(-1/2 (1 - x )^(-1/2))#

#= sqrt(1-x) - x/(2 sqrt(1-x))#

#= (2(1-x))/(2 sqrt(1-x)) - x/(2 sqrt(1-x))#

#= (2- 2x- x)/(2 sqrt(1-x))#

#= (2- 3x)/(2 sqrt(1-x))#