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

Question

Emma Aldridge · Accepted Answer

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

Charles Anctil · Answer

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

Isabella Ackerman · Answer

The answer is #=-3/(2sqrt(5-3x))# For this, we use #(sqrtu)'=1/(2sqrtu)# and the chain rule So, #(sqrt(5-3x))'=1/(2sqrt(5-3x))*(-3x)'# #=-3/(2sqrt(5-3x))#

Emilia Aguilar · Answer

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

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))#

Ezra Adams · Answer

undefined To find the derivative of $ \sqrt{5 - 3x} $, you can use the chain rule. First, rewrite the expression as $ (5 - 3x)^{\frac{1}{2}} $. Then, differentiate using the power rule and the chain rule. The derivative is:

$$ \frac{d}{dx} \left( \sqrt{5 - 3x} \right) = \frac{1}{2} (5 - 3x)^{-\frac{1}{2}} (-3) = \frac{-3}{2\sqrt{5 - 3x}} $$