# How do you find #(d^2y)/(dx^2)# given #y=(2x+5)^(-1/2)#?

##### 1 Answer

Nov 12, 2016

#### Explanation:

To find the

#color(blue)"first derivative" dy/dx# 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=2x+5rArr(du)/(dx)=2# and

#y=u^(-1/2)rArr(dy)/(du)=-1/2u^(-3/2)# substitute into (A) changing u back to x.

#rArrdy/dx=-1/2u^(-3/2)xx2=-(2x+5)^(-3/2)# To find the

#color(blue)"second derivative "(d^2y)/(dx^2) # repeat the process above on#dy/dx# using the#color(blue)"chain rule"#

#u=2x+5rArr(du)/(dx)=2#

#y=-u^(-3/2)rArr(dy)/(du)=3/2u^(-5/2)#

#rArr(d^2y)/(dx^2)=3(2x+5)^(-5/2)=3/(2x+5)^(5/2)#