If y = ea cos-1 x , show that from Mathematics Continuity and

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Continuity and Differentiability

Short Answer Type

571.

If space straight y equals left parenthesis cos to the power of negative 1 end exponent space straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis space straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0.

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572.

If space space space straight y equals left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x 2 right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0. space

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573.

If space space straight y equals open square brackets log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses close square brackets squared comma space prove space that space left parenthesis 1 space plus space straight x squared right parenthesis straight y subscript 2 space plus space straight x space straight y subscript 1 space equals space 2.

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574.

If space space straight y equals sin left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 plus straight m squared space straight y equals 0.

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575. If y = e^{m sin₁ x} , prove that (1 - x²) y₂ - x y₁ = m² y.

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576. If y = e^{a cos_{-1 x}} , show that
$left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight a squared straight y equals 0.$

space space straight y equals straight e to the power of straight m space sin to the power of negative 1 end exponent straight x end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis therefore space straight y subscript 1 equals space space straight y equals straight e to the power of straight m space sin to the power of negative 1 end exponent straight x end exponent. fraction numerator straight m over denominator square root of 1 minus straight x squared end root end fraction space space rightwards double arrow space space square root of 1 minus straight x squared end root. straight y subscript 1 equals straight m space straight e to the power of straight m space sin to the power of negative 1 end exponent straight x end exponent rightwards double arrow space square root of 1 minus straight x squared end root. straight y subscript 1 equals straight m space straight y space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket rightwards double arrow space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals straight m squared straight y squared

Differentiating both sides w.r.t. x, we get

(1-x²)·2 y₁ y₂+y₁² (-2 x)=a²·2 y y₁

Dividing both sides by 2 y₁ , we get,

left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 equals straight a squared straight y space space or space space left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight a squared straight y equals 0.

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577.

If space space straight f left parenthesis straight x right parenthesis equals vertical line straight x vertical line cubed comma space show space that space straight f " left parenthesis straight x right parenthesis space exists space for space all space real space straight x space and space find space it.

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578. If x = f(t), y = g (t) possess second order derivative for all t in (a, b) and f"(t) is inevitable then prove that

fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator straight f apostrophe left parenthesis straight t right parenthesis space straight g apostrophe apostrophe left parenthesis straight t right parenthesis minus straight g apostrophe left parenthesis straight t right parenthesis space straight f apostrophe apostrophe left parenthesis straight t right parenthesis over denominator left parenthesis straight f apostrophe left parenthesis straight t right parenthesis right parenthesis squared end fraction

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Long Answer Type

579.

If space left parenthesis straight x minus straight a right parenthesis squared space plus left parenthesis straight y minus straight b right parenthesis squared space equals straight c squared space comma space prove space that space space fraction numerator open square brackets 1 plus open parentheses begin display style dy over dx end style close parentheses squared close square brackets to the power of begin display style 3 over 2 end style end exponent over denominator begin display style fraction numerator straight d squared straight y over denominator dx squared end fraction end style end fraction space is space straight a space constant space independent space of space straight a space and space straight b.

86 Views

Short Answer Type

580.

Verify space Rolle apostrophe straight s space theorem space for space the space function straight f left parenthesis straight x right parenthesis equals fraction numerator 8 straight x squared over denominator 3 end fraction minus 2 straight x comma space straight x element of open square brackets 0 comma 3 over 4 close square brackets

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