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

Question

William Abdalla · Accepted Answer

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

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)#

Charles Arocha · Answer

See below

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))#

Adam Adkins · Answer

undefined To differentiate $ y = \sqrt{\frac{1}{2x^3 + 5}} $ using the chain rule:

1. Identify the outer function: $ \sqrt{x} $.
2. Identify the inner function: $ \frac{1}{2x^3 + 5} $.
3. Differentiate the outer function with respect to its argument: $ \frac{d}{du}(\sqrt{u}) = \frac{1}{2\sqrt{u}} $.
4. Differentiate the inner function with respect to $ x $: $ \frac{d}{dx}\left(\frac{1}{2x^3 + 5}\right) = -\frac{6x^2}{(2x^3 + 5)^2} $.
5. Apply the chain rule: Multiply the derivative of the outer function by the derivative of the inner function.

So, the derivative of $ y $ with respect to $ x $ is:

$$ y' = \frac{1}{2\sqrt{\frac{1}{2x^3 + 5}}} \cdot \left(-\frac{6x^2}{(2x^3 + 5)^2}\right) $$

Emilia Aguilar · Answer

undefined To use the chain rule to differentiate $ y = \sqrt{\frac{1}{2x^3 + 5}} $, we proceed as follows:

1. Let $ u = \frac{1}{2x^3 + 5} $.
2. Then $ y = \sqrt{u} $.
3. Now, differentiate $ u $ with respect to $ x $: $ \frac{du}{dx} = \frac{d}{dx}\left(\frac{1}{2x^3 + 5}\right) $.
4. Next, apply the chain rule to differentiate $ y $ with respect to $ x $: $ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $.
5. Finally, substitute $ \frac{du}{dx} $ and $ u $ back into the expression to find $ \frac{dy}{dx} $.