# How do you use the chain Rule to find the derivative of sqrt(2x^3 - 3x- 4)?

Jul 29, 2015

${y}^{'} = \frac{3}{2} \cdot \frac{2 {x}^{2} - 1}{\sqrt{2 {x}^{3} - 3 x - 4}}$

#### Explanation:

You can use the chain rule to find the derivative of this function by using $u = 2 {x}^{3} - 3 x - 4$.

The chain rule tells you that you can differentiate a function $y$ that depends on a variable $u$, which in turn depends on another variable $x$, by

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{du}} \left(y\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)}$

Two additional rules of derivation, the power rule and the sum rule will come into play at certain points in the derivation of $y$.

So, if you use $u = 2 {x}^{3} - 3 x - 4$, then you have

$y = \sqrt{u} = {u}^{\frac{1}{2}}$

So, the derivative of $y$ will look like this

${y}^{'} = \frac{d}{\mathrm{du}} \left(y\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

${y}^{'} = \frac{d}{\mathrm{du}} \left({u}^{\frac{1}{2}}\right) \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 3 x - 4\right)$

The power rule tells you that you can differentiate a variable $x$ raised to a power $a$ by

$\textcolor{b l u e}{{x}^{a} = a \cdot {x}^{a - 1}}$

This means that you have

$\frac{d}{\mathrm{du}} \left({u}^{\frac{1}{2}}\right) = \frac{1}{2} \cdot {u}^{\frac{1}{2} - 1}$

$\frac{d}{\mathrm{du}} \left({u}^{\frac{1}{2}}\right) = \frac{1}{2} \cdot \frac{1}{u} ^ \left(\frac{1}{2}\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{u}}$

The sum rule tells you that the derivative of a function that can be written as a sum of two (or more) functions is equal to the sum of the derivatives of those functions.

For $\textcolor{b l u e}{y = f \left(x\right) + g \left(x\right) + h \left(x\right) + \ldots}$

you have

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = {f}^{'} \left(x\right) + {g}^{'} \left(x\right) + {h}^{'} \left(x\right) + \ldots}$

This means that you can write

$\frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 3 x - 4\right) = \frac{d}{\mathrm{dx}} \left(2 {x}^{3}\right) + \frac{d}{\mathrm{dx}} \left(- 3 x\right) + \frac{d}{\mathrm{dx}} \left(- 4\right)$

$\frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 3 x - 4\right) = 2 \cdot 3 {x}^{3 - 2} - 3 + 0$

$\frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 3 x - 4\right) = 6 {x}^{2} - 3$

Your original derivative now becomes

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

y^] = 1/2 * 1/sqrt((2x^3 - 3x - 4)) * 3(2x^2 - 1)

Finally,

${y}^{'} = \textcolor{g r e e n}{\frac{3}{2} \cdot \frac{2 {x}^{2} - 1}{\sqrt{2 {x}^{3} - 3 x - 4}}}$