How do you use the chain rule to differentiate #y=2(x+3)^(1/2)#?

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Answer 1 · 2018-06-05T18:04:07

#dy/dx=(x+3)^(-1/2)#

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

Let #u(x)=x+3#

Differentiating this with powers rule:

#(du)/dx=1#

This means #y(u)=2u^(1/2)#

Differentiate this with the powers rule.

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

Now, we can use the fact that

#dy/dx=dy/(du) *(du)/dx#

#dy/dx=u^(-1/2)*1#

Since #u=x+3#

#dy/dx=(x+3)^(-1/2)#