# How do you differentiate f(x)=sec(1/x^3) using the chain rule?

##### 1 Answer
Nov 29, 2016

$f ' \left(x\right) = - \frac{3}{x} ^ 4 \sec \left(\frac{1}{x} ^ 3\right) \tan \left(\frac{1}{x} ^ 3\right)$

#### Explanation:

Differentiating this function is determined by applying chain rule.
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Chain Rule:
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$\textcolor{b l u e}{\left(h \left(g \left(x\right)\right)\right) ' = h ' \left(g \left(x\right)\right) \times g ' \left(x\right)}$
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Applying the chain rule :
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$\left(\sec \left(\frac{1}{x} ^ 3\right)\right) ' = \sec ' \left(\frac{1}{x} ^ 3\right) \times \left(\frac{1}{x} ^ 3\right) '$
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Knowing that:
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$\frac{d}{\mathrm{dx}} \left(\sec x\right) = \tan x \sec x$
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$\sec ' \left(\frac{1}{x} ^ 3\right) = \tan \left(\frac{1}{x} ^ 3\right) \times \sec \left(\frac{1}{x} ^ 3\right)$
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$\frac{1}{x} ^ 3 \text{ }$ can be differentiated by either applying quotient or power rule, here I will apply the power rule.
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$\left(\frac{1}{x} ^ 3\right) ' = {x}^{- 3} = - 3 {x}^{- 4} = - \frac{3}{x} ^ 4$
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Therefore,
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$\left(\sec \left(\frac{1}{x} ^ 3\right)\right) ' = \sec ' \left(\frac{1}{x} ^ 3\right) \times \left(\frac{1}{x} ^ 3\right) '$
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$\left(\sec \left(\frac{1}{x} ^ 3\right)\right) ' = \sec \left(\frac{1}{x} ^ 3\right) \times \tan \left(\frac{1}{x} ^ 3\right) \times \left(- \frac{3}{x} ^ 4\right)$
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Hence,
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$f ' \left(x\right) = - \frac{3}{x} ^ 4 \sec \left(\frac{1}{x} ^ 3\right) \tan \left(\frac{1}{x} ^ 3\right)$