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यदि (If) `y=log(1+cosx)`, सिद्ध करें कि (prove that) `(d^(3)y)/(dx^(3))+(d^(2)y)/(dx^(2)).(dy)/(dx)=0` |
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Answer» दिया है, `y=log(1+cosx)" "...(1)` `implies" "(dy)/(dx)=(1)/(1+cosx)(-sinx)=(-sinx)/(1+cosx)" "...(2)` `implies" "(d^(2)y)/(dx^(2))=-((1+cosx)cdot cosx-sinx(-sinx))/((1+cosx)^(2))` `=-(cosx+cos^(2)x+sin^(2)x)/((1+cosx)^(2))` `=-(cosx+1)/((1+cosx)^(2))=(-1)/(1+cosx)" "...(3)` `implies" "(d^(3)y)/(dx^(3))=(-1)(-1)(1+cosx)^(-2)(-sinx)` `=(-sinx)/((1+cosx)^(2))" "...(4)` अब `(d^(3)y)/(dx^(3))+(d^(2)y)/(dx^(2)).(dy)/(dx)` `=(-sinx)/((1+cosx)^(2))+(1)/(1+cosx)xx(-sinx)/(1+cosx)" "[(2), (3)" और "(4)" से"]` `=(-sinx+sinx)/((1+cosx)^(2))=0` |
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