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Could u help me solve the derivative of inverse trignometric problem?

y = 2 tan^(-1) 3x, find dy/dx

..........................................

y/2 = tan^(-1) 3x

tan y/2 = 3x

sec^(2) y/2 . dy/dx = 3

dy/dx = 3/(sex^(2) y/2)

dy/dx = 3/(1+tan^(2) y/2)

dy/dx = 3/(1+(3x)^2)

dy/dx = 3/(1+9x^2) <=== (Check my answer pls)

2 Answers

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  • 1 decade ago
    Favorite Answer

    When you took the x derivative of tan (y/2), you got (sec(y/2))^2 dy/dx, but because of the chain rule, you would get an additional factor of 1/2, from the derivative of the inside. That 1/2 would end up in the denominator of your final answer resulting in an overall factor of 2 change, so

    dy/dx = 6/(1+9x^2)

  • Ed I
    Lv 7
    1 decade ago

    y = 2 tan^(-1) 3x

    dy/dx = 2 • 1/(1 + (3x)^2) •Â 3 = 6/(1 + 9x^2)

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