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Math Help - Trigo derivative urgent help please...

  1. #1
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    Trigo derivative urgent help please...

    what is the derivative of (sinx)^tanx??
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  2. #2
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    logarithmic differentiation ...


    y = (\sin{x})^{\tan{x}}

    \ln(y) = \ln\left[(\sin{x})^{\tan{x}}\right]

    \ln(y) = \tan{x} \cdot \ln(\sin{x})

    take the derivative w/r to x ...

    \frac{y'}{y} = \tan{x} \cdot \frac{\cos{x}}{\sin{x}} + \ln(\sin{x}) \cdot \sec^2{x}

    \frac{y'}{y} = 1 + \sec^2{x} \cdot \ln(\sin{x})

    y' = (\sin{x})^{\tan{x}}\left[1 + \sec^2{x} \cdot \ln(\sin{x})\right]
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by juneau View Post
    what is the derivative of (sinx)^tanx??
    sin(x)^{tan(x)} = y

    log(y) = tan(x)log(sin(x))

    <br />
\frac{1}{y} \frac{dy}{dx} = sec^2(x)log(sin(x)) + tan(x)\times cos(x)\times \frac{1}{sin(x)}  ------Using Chain Rule

    <br />
  \frac{dy}{dx} = y\times (sec^2(x)log(sin(x)) + 1)

    If you want put the value of y in RHS that's your answer
    EDIT: Skeeter won
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  4. #4
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    well see, the original question was: y = (x)^(sinx)^(tanx)

    I already used log differentiation to get (sinx)^tanx down. can I use log differentiation again??
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  5. #5
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    Quote Originally Posted by juneau View Post
    well see, the original question was: y = (x)^(sinx)^(tanx)

    I already used log differentiation to get (sinx)^tanx down. can I use log differentiation again??
    why did you not mention that to begin with?

    y = x^{\sin{x}^{\tan{x}}}

    \ln{y} = \ln\left(x^{\sin{x}^{\tan{x}}}\right)

    \ln{y} = \sin{x}^{\tan{x}} \cdot \ln{x}

    use the product rule ...

    \frac{y'}{y} = \sin{x}^{\tan{x}} \cdot \frac{1}{x} + \ln{x} \cdot (derivative \, from \, my \, previous \, post)

    finish up.
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  6. #6
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    Yes, you can, using the chain rule.


    By the way, here's an interesting point: if y= f(x)^{g(x)}, where f and g are integrable functions, there are two common mistakes in finding the derivative:

    1) Ignore the fact that g is a variable and use the power law:
    y'= g(x) f(x)^{g(x)-1}

    2) Ignore the fact that f is a variable and treat it as an exponential:
    y'= ln(f(x)) f(x)^{g(x)}

    As I said, those are both wrong. The correct derivative is the sum of those!

    y'= g(x) f(x)^{g(x)-1}+ ln(f(x)) f(x)^{g(x)}
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