How do you find the derivative of #sqrt(5-3x)#?

4 Answers
Nov 5, 2016

#=(-3)/(2sqrt(5-3x))#

Explanation:

To find the derivative of the expression is by applying chain rule
Let #v(x)=5-3x and u(x)=sqrtx#

Then #u@v(x)=u(v(x))=sqrt(5-3x)#

Then #color(blue)((sqrt(5-3x))'=(u@v(x))'=u'(v(x))xxv'(x)#

Computing #u'(v(x)) and v'(x)#

#u'(x)=1/(2sqrtx) #
Then

#color(blue)( u'(v(x))=1/(2sqrt(5-3x))#

#color(blue)(v'(x)=-3#

#color(blue)((sqrt(5-3x))'=(u@v(x))'=u'(v(x))xxv'(x)#

#(sqrt(5-3x))'=1/(2sqrt(5-3x))xx-3#

Therefore,
#(sqrt(5-3x))'=(-3)/(2sqrt(5-3x))#

Nov 5, 2016

# d/dx sqrt(5-3x) = -3/(2sqrt(5-3x) #

Explanation:

# f(x)= sqrt(5-3x) #
We know,
#d/dx(sqrtx) = 1/(2sqrtx)#
and also,
# d/dx(a-bx)=-b#

hence, using the chain rule, we differentiate #f(x)# to get
# d/dx sqrt(5-3x) = -3/(2sqrt(5-3x) #

Nov 5, 2016

The answer is #=-3/(2sqrt(5-3x))#

Explanation:

For this, we use #(sqrtu)'=1/(2sqrtu)# and the chain rule

So, #(sqrt(5-3x))'=1/(2sqrt(5-3x))*(-3x)'#

#=-3/(2sqrt(5-3x))#

Nov 5, 2016

#-3/(2sqrt(5-3x))#

Explanation:

Express #y=sqrt(5-3x)=(5-3x)^(1/2)#

differentiate using the #color(blue)"chain rule"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))to(A)#

let #u=5-3xrArr(du)/(dx)=-3#

and #y=u^(1/2)rArr(dy)/(du)=1/2u^(-1/2)#

substitute these values into (A) changing u back into terms of x.

#rArrdy/dx=1/2u^(-1/2)xx(-3)=-3/(2u^(1/2))=-3/(2sqrt(5-3x))#