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`e^(6x)"cos"3x` |
Answer» Let `y=e^(6x)" cos"3x` `implies(dy)/(dx)=(d)/(dx)(e^(6x)"cos"3x)` `=e^(6x)*(-3" sin"3x)+"cos"3x*(6e^(6x))` `=e^(6x)(-3" sin"3x+6" cos"3x)` `implies (d^(2)y)/(dx^(2))=(d)/(dx)[e^(6x)(-3" sin"3x+6" cos"3x)]` `=e^(6x)(-9" cos"3x-18" sin"3x)+(-3" sin"3x+6" cos"3x)*(6e^(6x))` `=e^(6x)(-9" cos"3x-18" sin"3x-18" sin"3x+36" cos"3x)` `=e^(6x)(27" cos"3x-36" sin"3x)` `=9e^(6x)(3" cos"3x-4" sin"3x)` |
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