How do you differentiate #y=sqrt(4+3x)#?

1 Answer
Feb 27, 2017

#y=3/(2(4+3x)^(1/2))#

Explanation:

Recall the formula for chain rule:

#color(blue)(bar(ul(|color(white)(a/a)dy/dx=dy/(du)(du)/dxcolor(white)(a/a)|)))# or #color(blue)(bar(ul(|color(white)(a/a)f'(x)=g'[h(x)]h'(x)color(white)(a/a)|)))#

and the formula for power rule:

#color(blue)(bar(ul(|color(white)(a/a)d/dx(x^n)=nx^(n-1)color(white)(a/a)|)))#

To start, recognize the inside and outside functions of #y=color(green)(sqrt(color(darkorange)(4+3x)))#.

Inside function: #y=color(darkorange)(4+3x)#
Outside function: #y=color(green)(sqrt(a))#

How to Differentiate Using Chain Rule

#1#. Take the derivative of the outside function, #y=sqrt(a)#, but replace the #a# with the inside function, #4+3x#.

#2#. Multiply by the derivative of the inside function, #4+3x#.

Applying Chain Rule
1. The derivative of the outside function, #y=sqrt(a)#, would be #1/2a^(-1/2)#, using the power rule. However, #a# needs to be replaced by the inside function, so it becomes #1/2(4+3x)^(-1/2)#.

#y=sqrt(a)#

#color(red)(darr)#

#y=1/2a^(-1/2)#

#color(red)(darr)#

#y=1/2(4+3x)^(-1/2)#

2.#color(white)(i)#Then we need to multiply by the derivative of the inside function, #4+3x#, which becomes #3# using the power rule.

#y=1/2(4+3x)^(-1/2)#

#color(red)(darr)#

#y=1/2(4+3x)^(-1/2)(3)#

#color(green)(bar(ul(|color(white)(a/a)y=3/(2(4+3x)^(1/2))color(white)(a/a)|)))#