 zatajuxoqj

2022-03-22

What is the equation of the tangent line of $f\left(x\right)=\frac{{\mathrm{cos}}^{2}x}{{e}^{x}}-x\mathrm{tan}x$ at $x=\frac{\pi }{4}$? shvatismop1rj

First, find the point the tangent line will intersect. I'll be using decimals since there is no clean way to simplify this by any means.
$f\left(\frac{\pi }{4}\right)=-0.5574$
The tangent line will intersect the point
(Replacing $\frac{\pi }{4}$ with $0.7854.$)
To find the slope of the tangent line, find ${f}^{\prime }\left(\frac{\pi }{4}\right)$.
To find f'(x), use quotient rule for the first term and product rule for $x\mathrm{tan}x$.
${f}^{\prime }\left(x\right)=\frac{2\mathrm{cos}x\left(-\mathrm{sin}x\right){e}^{x}-{e}^{x}{\mathrm{cos}}^{2}x}{{e}^{2x}}-\mathrm{tan}x-x{\mathrm{sec}}^{2}x$
$=\frac{-\mathrm{cos}x\left(2\mathrm{sin}x+\mathrm{cos}x\right)}{{e}^{x}}-\mathrm{tan}x-x{\mathrm{sec}}^{2}x$
Plug in $\frac{\pi }{4}$.
${f}^{\prime }\left(\frac{\pi }{4}\right)=-3.2547$
Plug the point and slope into an equation in point-slope form:
$y+0.5574=-3.2547\left(x-0.7854\right)$

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