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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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