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

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 $u$ :

$\frac{d}{\mathrm{du}} \sqrt{u} = \frac{1}{2 \sqrt{u}}$

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, $\frac{1}{2 {x}^{3} + 5}$, let's differentiate with respect to $x$ using the quotient rule:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{2 {x}^{3} + 5}\right) = \frac{\left(2 {x}^{3} + 5\right) \left(0\right) - \left(1\right) \left(6 {x}^{2}\right)}{2 {x}^{3} + 5} ^ 2$

$\frac{d}{\mathrm{dx}} \left(\frac{1}{2 {x}^{3} + 5}\right) = - \frac{6 {x}^{2}}{2 {x}^{3} + 5} ^ 2$

Let's multiply our differentiated layers together:

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

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 $y = \sqrt{\frac{1}{2 {x}^{3} + 5}}$ can be re-written:

$y = {\left(\frac{1}{2 {x}^{3} + 5}\right)}^{\frac{1}{2}}$

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

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

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

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

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.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} {\left(2 {x}^{3} + 5\right)}^{- \frac{3}{2}} \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{3} + 5\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} {\left(2 {x}^{3} + 5\right)}^{- \frac{3}{2}} \cdot 6 {x}^{2}$

Simplifying and converting negative exponents to positive exponents gives us:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {x}^{2}}{2 {x}^{3} + 5} ^ \left(\frac{3}{2}\right) = \frac{- 3 {x}^{2}}{\sqrt{{\left(2 {x}^{3} + 5\right)}^{3}}}$