Question

Find the equation of the tangent to the curve with equation
$y=3\cos h2x-\sin hx$ at the point $x=\ln 2$

Answer 1

Slope of the curve at $x=ln2$ is $y^{{}^{\prime }}\left(ln2\right)$

$\Rightarrow y^{{}^{\prime }}\left(x\right)=6sinh2x-coshx$

$\Rightarrow y^{{}^{\prime }}\left(ln2\right)=6\left(\frac{e^{ln2}-e^{-ln2}}{2}\right)-\left(\frac{e^{ln2}+e^{-ln2}}{2}\right)$

$\Rightarrow y^{{}^{\prime }}\left(ln2\right)=6\left(2\right)-0$

$\Rightarrow y^{{}^{\prime }}\left(ln2\right)=12$

And $y\left(ln2\right)=-2$

Equation of tangent =$y-(-2)=12(x-ln2)$

$\Rightarrow y=12x-(2+12ln2)$ is the equation of the tangent