Question

# Differentiate $secx$ by the first principle.

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Solution

## Let us consider$secx=\frac{1}{\mathrm{cos}x}$By quotient rule derivative of $secx=\frac{1}{\mathrm{cos}x}$[The quotient rule of differentiation is defined as the ratio of two functions (1st function / 2nd function ), is equal to the ratio of (differentiation of 1st function $×$the 2nd function $-$ differentiation of the second function $×$the first function ) to the square of the 2nd function]so,$\frac{\frac{d}{dx}×1-\frac{d}{dx}\mathrm{cos}x×1}{{\left(\mathrm{cos}x\right)}^{2}}\phantom{\rule{0ex}{0ex}}$by simplifying the equation$=\frac{0×\mathrm{cos}x-\frac{d}{dx}\mathrm{cos}x×1}{{\left(\mathrm{cos}x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}x}{\mathrm{cos}x×\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}x}{\mathrm{cos}x}×\frac{1}{\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=tanx×secx$Thus, differentiation of $secx$ by the first principle is$\mathbf{}\mathbit{t}\mathbit{a}\mathbit{n}\mathbit{x}\mathbf{×}\mathbit{s}\mathbit{e}\mathbit{c}\mathbit{x}$.

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