Question

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

Solution

## Slope of the curve at $$x=ln2$$ is $$y^{'} (ln2)$$$$\Rightarrow y^{'} (x) =6sinh2x-coshx$$$$\Rightarrow y^{'} (ln2) =6(\dfrac {e^{ln2} - e^{-ln2}} {2})-(\dfrac {e^{ln2} +e^{-ln2}} {2})$$$$\Rightarrow y^{'} (ln2) =6(2)-0$$$$\Rightarrow y^{'} (ln2) =12$$And $$y(ln2) =-2$$Equation of tangent =$$y-(-2)=12(x-ln2)$$$$\Rightarrow y=12x-(2+12ln2)$$ is the equation of the tangent Mathematics

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