Question

How to Derive the derivative of secx

Answer 1

Dear student,
$$\begin{gathered} Lety=secx \\ \Rightarrow y=\frac{1}{\cos x} \\ ondifferentiating \\ \Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{\cos x}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{\frac{d}{dx}\left(1\right).\cos x-1.\frac{d}{dx}\left(\cos x\right)}{{\left(\cos x\right)}^2}--[u\sin gquotientrule] \\ \Rightarrow \frac{dy}{dx}=\frac{\left(0\right)\left(\cos x\right)-\left(1\right)(-\sin x)}{{\cos }^{2}x} \\ \Rightarrow \frac{dy}{dx}=\frac{\sin x}{{\cos }^{2}x} \\ \Rightarrow \frac{dy}{dx}=\frac{\sin x}{\cos x}.\frac{1}{\cos x} \\ \Rightarrow \frac{dy}{dx}=\tan x.secx \end{gathered}$$
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