class 12 maths continuity and differentiability

${e^{6x}}\cos 3x$

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📘 Continuity and Differentiability NCERT Ex.5.7 ,Q.7,Page 183 SA

${e^{6x}}\cos 3x$

Official Solution

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Let$y = {e^{6x}}\cos 3x$
$\Rightarrow$ $\cfrac{{dy}}{{dx}} = {e^{6x}}( - \sin 3x) \cdot 3 + \cos 3x \cdot {e^{6x}} \cdot 6 = 6{e^{6x}}\cos 3x - 3{e^{6x}}\sin 3x$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} = 6[{e^{6x}}( - \sin 3x) \cdot 3 + \cos 3x \cdot {e^{6x}}.6]$ $- 3[{e^{6x}}\cos 3x \cdot 3 + \sin 3x{e^{6x}} \cdot 6]$

$= - 18{e^{6x}}\sin 3x + 36{e^{6x}}\cos 3x - 9{e^{6x}}\cos 3x - 18{e^{6x}}\sin 3x$
$= 27{e^{6x}}\cos 3x - 36{e^{6x}}\sin 3x = 9{e^{6x}}(3\cos 3x - 4\sin 3x)$

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