# Chain Rule

Differentiate x^(lnx) using implicit differentiation and the chain rule

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 10 videos by Tiago Hands

## Key Questions

• The Chain Rule is the idea that the derivative of a function can be expanded into two component parts, namely that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dy}}{\mathrm{du}}$

In the chain rule, another variable (this time $u$) is introduced and used to expand $\frac{\mathrm{dy}}{\mathrm{dx}}$. Because $\mathrm{du}$ will be cancelled out during multiplication the equation works.

This can be used to find the differential of any function with a bracket raised to a power.

For example: If $y = {\left(5 x + 3\right)}^{4}$ find $\frac{\mathrm{dy}}{\mathrm{dx}}$:

Let $u = 5 x + 3$ and $y = {u}^{4}$

$\frac{\mathrm{du}}{\mathrm{dx}} = 5$ and $\frac{\mathrm{dy}}{\mathrm{du}} = 4 {u}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dy}}{\mathrm{du}}$

$= 4 {u}^{3} \cdot 5$

$= 20 {u}^{3}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 20 {\left(5 x + 3\right)}^{3}$ substituting back for u

Another simple way to do the same thing is to use The Power Rule. The rule states "Bring down the power, reduce the power of the brackets by one, and multiply by the derivative of the function within the brackets".

To break it down for the same example of $y = {\left(5 x + 3\right)}^{4}$ we will go through it step by step:

1. "Bring down the power": Here we write the original power from the brackets $y ' = 4$
2. "Reduce the power of the brackets by one": $y ' = 4 {\left(5 x + 3\right)}^{3}$
3. "Multiply by the derivative of the function within the brackets": $y ' = 4 {\left(5 x + 3\right)}^{3} \times 5$
4. $\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 20 {\left(5 x + 3\right)}^{3}$

Both of these are sometimes known as the chain rule but you should know both variants.

• We can do a number of things ..

1) Use the chain rule and quotient rule

2) Use the chain rule and the power rule after the following transformations.

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

3) You could multiply out everything, which takes a bunch of time, and then just use the quotient rule.

Let's keep it simple and just use the chain rule and quotient rule.

Chain Rule $\implies y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \left(\frac{1 + x}{1 - x}\right) '$

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

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

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

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

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

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

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

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

$y ' = 6 \frac{{\left(1 + x\right)}^{2}}{1 - x} ^ 4$

• By Chain Rule and Power Rule,

$y ' = \frac{1}{4} {\left({x}^{2} + 3 x + 5\right)}^{\frac{1}{4} - 1} \cdot \left({x}^{2} + 3 x + 5\right) '$

$= \frac{1}{4} {\left({x}^{2} + 3 x + 5\right)}^{- \frac{3}{4}} \left(2 x + 3\right)$

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