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anonymous

  • one year ago

Solve, If f'(x) = Cos(x^2) and y = f [ (2x-1) / (x+1) ], find dy/dx

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  1. ganeshie8
    • one year ago
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    we may use chain rule \[y=f(\color{blue}{g(x)}) ~\implies ~\dfrac{dy}{dx}=f'(\color{blue}{g(x)})\times \color{blue}{g'(x)}\]

  2. anonymous
    • one year ago
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    Okay, I ended up having dy/dx = f' [ (2x-1) / (x+1) ] x [ 3 / (x+1)^2 ]

  3. ganeshie8
    • one year ago
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    looks good, plugin the value of f' [ (2x-1) / (x+1) ]

  4. ganeshie8
    • one year ago
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    f'(x) = cos(x^2) f' [ (2x-1) / (x+1) ] = ?

  5. anonymous
    • one year ago
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    Would it equal Cos(x^2)? Or do you have to substitute Cos(x^2) into f'?

  6. anonymous
    • one year ago
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    Like as in f' [ (2x-1) / (x+1) ] = Cos(x^2) OR Cos(x^2) [ (2x-1) / (x+1) ]

  7. ganeshie8
    • one year ago
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    not quite, simply replace x by (2x-1) / (x+1)

  8. anonymous
    • one year ago
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    Ah okay So like: Cos( [ (2x-1) / (x+1) ] ^2 ) ?

  9. ganeshie8
    • one year ago
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    \[f'(\color{red}{x}) = \cos(\color{red}{x}^2)\] so \[f'(\color{red}{ (2x-1) / (x+1) }) = \cos(\color{red}{ [ (2x-1) / (x+1) ] }^2)\]

  10. anonymous
    • one year ago
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    Ah okay thank you ^^

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