Write the equation of the line tangent to f(x) = tan x + 3 at (pi/4. 4)

opatovaL 2021-09-10 Answered
Write the equation of the line tangent to \(\displaystyle{f{{\left({x}\right)}}}={\tan{{x}}}+{3}\) at \(\displaystyle{\left(\frac{\pi}{{4}}.{4}\right)}\)

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Expert Answer

ottcomn
Answered 2021-09-11 Author has 12767 answers
\(\displaystyle{f{{\left({x}\right)}}}={\tan{{x}}}+{3}\) at \(\displaystyle{\left(\frac{\pi}{{4}}.{4}\right)}\)
\(\displaystyle{y}={f{{\left({x}\right)}}}={\tan{+}}{3}\)
\(\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}\times{f{{\left({x}\right)}}}=\frac{{d}}{{\left.{d}{x}\right.}}{\left({\tan{{x}}}+{3}\right)}\)
\(\displaystyle{f}'{\left({x}\right)}={{\sec}^{{2}}{x}}+{0}\)
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={{\sec}^{{2}}{x}}\)
Slope \(\displaystyle{m}=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={{\sec}^{{2}}\times}\frac{\pi}{{4}}={{\sec}^{{2}}{\left(\frac{\pi}{{4}}\right)}}={\left(\sqrt{{2}}\right)}^{{2}}={2}\)
\(\displaystyle{\left({y}–{4}\right)}={2}{\left({x}–\frac{\pi}{{x}}\right)}\)
\(\displaystyle{y}={2}{x}–\frac{\pi}{{2}}+{4}\)
\(\displaystyle{y}={2}{x}–\frac{{\pi+{8}}}{{2}}\)
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