Question

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

Answer 1

`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)`

Answer 2

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))`