Question

Set up an equation of a tangent to the graph of the following function.
$y=2^{-x}-2^{-2x}$ at the point with abscissa $x=2.$

Answer 1

If $x_1=2$ ,$y_1=\frac{3}{16}$

$\frac{dy}{dx}=({2.2}^{-2x}-2^{-x})\ln 2$

${\left(\frac{dy}{dx}\right)}_{x=2}=\frac{1}{8}-\frac{1}{4}=-\frac{1}{8}\ln 2=m$

$(x-x_1)m=y-y_1$

$(x-2)(-\frac{1}{8}\ln 2)=y-\frac{3}{16}$

$\left(2\ln 2\right)x+16y-3-4\ln 2=0$ is the tangent at the point with abscissa ${}^{\prime }{2}^{\prime }$