Question

if f(x)=log(secx+tanx) then find f(x)

Answer 1

$\textbf{Given:}$

$\mathsf{f(x)=\log(secx+tanx)}$

$\textbf{To find:}$

$\mathsf{f'(x)}$

$\textbf{Solution:}$

$\underline{\textbf{Formula used:}}$

$\boxed{\begin{minipage}{4cm} \\\\\mathsf{\dfrac{d(logx)}{dx}=\dfrac{1}{x}}\\\\\mathsf{\dfrac{d(secx)}{dx}=secx\,tanx}\\\\\mathsf{\dfrac{d(tanx)}{dx}=sec^2x}\\ \end{minipage}}$

$\textsf{We apply Chain rule to differentiate the given function}$

$\mathsf{Consider,}$

$\mathsf{f(x)=\log(secx+tanx)}$

$\textsf{Differentiate with respect to 'x'}$

$\mathsf{f\,'(x)=\dfrac{1}{secx+tanx}{\times}\dfrac{d(secx+tanx)}{dx}}$

$\mathsf{f\,'(x)=\dfrac{1}{secx+tanx}{\times}(secx\,tanx+sec^2x)}$

$\mathsf{f\,'(x)=\dfrac{1}{secx+tanx}{\times}(tanx+secx)\,secx}$

$\mathsf{f\,'(x)=\dfrac{1}{secx+tanx}{\times}(secx+tanx)\,secx}$

$\implies\boxed{\mathsf{f\,'(x)=secx}}$

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