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

Write the equation of the line tangent to $$\displaystyle{f{{\left({x}\right)}}}={\tan{{x}}}+{3}$$ at $$\displaystyle{\left(\frac{\pi}{{4}}.{4}\right)}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

ottcomn
$$\displaystyle{f{{\left({x}\right)}}}={\tan{{x}}}+{3}$$ 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 $$\displaystyle{m}=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={{\sec}^{{2}}\times}\frac{\pi}{{4}}={{\sec}^{{2}}{\left(\frac{\pi}{{4}}\right)}}={\left(\sqrt{{2}}\right)}^{{2}}={2}$$
$$\displaystyle{\left({y}–{4}\right)}={2}{\left({x}–\frac{\pi}{{x}}\right)}$$
$$\displaystyle{y}={2}{x}–\frac{\pi}{{2}}+{4}$$
$$\displaystyle{y}={2}{x}–\frac{{\pi+{8}}}{{2}}$$