How do you use the chain rule to differentiate #y=sqrt(1/(2x^3+5))#?

2 Answers
Feb 23, 2018

#dy/dx=-(3x^2)/((2x^3+5)^(3/2)#

Explanation:

When differentiating functions with the chain rule, it helps to think of our function as "layered," remembering that we must differentiate one layer at a time, from the outermost layer to the innermost layer, and multiply these results.

Here, our outer layer would be the square root, while the inner layer would be the quotient of a polynomial.

Let's temporarily denote everything inside the root by #u# and differentiate our outer layer with respect to #u# :

#(d)/(du)sqrt(u)=1/(2sqrt(u))#

Let's rewrite this in terms of #x:#

#1/(2sqrt((1/(2x^3+5)))#

We've differentiated our outer layer. Moving on to the inner layer, #1/(2x^3+5)#, let's differentiate with respect to #x# using the quotient rule:

#d/dx(1/(2x^3+5))=((2x^3+5)(0)-(1)(6x^2))/(2x^3+5)^2#

#d/dx(1/(2x^3+5))=-(6x^2)/(2x^3+5)^2#

Let's multiply our differentiated layers together:

#dy/dx=1/(2sqrt((1/(2x^3+5))))*-(6x^2)/(2x^3+5)^2#

Simplify:

#dy/dx=-(3x^2)/((2x^3+5)^2/sqrt(2x^3+5)#

#dy/dx=-(3x^2)/((2x^3+5)^2/(2x^3+5)^(1/2)#

Using the division rule for exponents, we get:
#dy/dx=-(3x^2)/((2x^3+5)^(3/2)#

Feb 23, 2018

See below

Explanation:

So, for square roots and other nth-root functions, I personally always convert them to rational exponents. There are probably other ways to do it, but the students I help always seem to like this method, too.

So the function #y=sqrt(1/(2x^3+5))# can be re-written:

#y=(1/(2x^3+5))^(1/2)#

I'll even go one further by changing the inside to a negative exponent so I can avoid using a quotient rule:

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

This leads us to something interesting. We can actually multiply the exponents (taking a power to a power), so now we have:

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

So here's where we use the chain rule for differentiation. Basically, we take the derivative of the outside-most function multiplied by the derivative of the inside function. I like to do this in steps so I don't get confused.

#dy/dx=-1/2(2x^3+5)^(-3/2)*d/dx(2x^3+5)#

#dy/dx=-1/2(2x^3+5)^(-3/2)*6x^2#

Simplifying and converting negative exponents to positive exponents gives us:

#dy/dx=(-3x^2)/(2x^3+5)^(3/2)=(-3x^2)/sqrt((2x^3+5)^(3))#