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tan^(-1)((7x)/(1-12x^(2)))...

`tan^(-1)((7x)/(1-12x^(2)))`

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To solve the problem of differentiating \( y = \tan^{-1}\left(\frac{7x}{1 - 12x^2}\right) \), we will follow these steps: ### Step 1: Recognize the Identity We can use the identity for the tangent inverse function: \[ \tan^{-1}\left(\frac{a + b}{1 - ab}\right) = \tan^{-1}(a) + \tan^{-1}(b) \] In our case, we can express \( \frac{7x}{1 - 12x^2} \) in the form of this identity by letting \( a = 4x \) and \( b = 3x \). ...
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If 2sin^(-1)((9-4x^(2))/(9+4x^(2)))+3cos^(-1)((12x)/(9+4x^(2)))+tan^(-1)((12x)/(9-4x^(2)))=lambda(pi)/(2) for x=(-17)/(13) "then" lambda

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Knowledge Check

  • 5cos^(-1)((1-x^(2))/(1+x^(2)))+7sin^(-1)((2x)/(1+x^(2)))-4tan^(-1)((2x)/(1-x^(2)))-tan^(-1)x=5pi , then x is equal to

    A
    3
    B
    `-sqrt(3)`
    C
    `sqrt(2)`
    D
    `sqrt(3)`
  • (d)/(dx)[tan^(-1)((6x)/(1+7x^(2)))]+(d)/(dx)[tan^(-1)((5+2x)/(2-5x))]=

    A
    `(7)/(1+49x^(2))`
    B
    `(2)/(1+4x^(2))`
    C
    `(-5)/(1+25x^(2))`
    D
    `(9)/(1+3x^(2))`
  • If 5 cos^(-1)((1-x^2)/(1+x^2))+7sin^(-1)((2x)/(1+x^2))-4tan^(-1)((2x)/(1-x^2))-tan^(-1)x=5pi , then x is equal to

    A
    3
    B
    `-sqrt3`
    C
    `sqrt2`
    D
    `sqrt3`
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