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# 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

$f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

#### Explanation:

In differential calculus, we use the Chain Rule when we have a composite function. It states:

The derivative will be equal to the derivative of the outside function with respect to the inside, times the derivative of the inside function. Let's see what that looks like mathematically:

Chain Rule:

$f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Let's say we have the composite function $\sin \left(5 x\right)$. We know:

$f \left(x\right) = \sin x \implies f ' \left(x\right) = \cos x$

$g \left(x\right) = 5 x \implies g ' \left(x\right) = 5$

So the derivative will be equal to

$\cos \left(5 x\right) \cdot 5$

$= 5 \cos \left(5 x\right)$

We just have to find our two functions, find their derivatives and input into the Chain Rule expression.

Hope this helps!

• 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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