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Find ${\dfrac {dy}{dx}}$

$$y=(\ln x)^x+(\sin^{-1}x)^{\sin x} , \quad x\in \left(0,\;\frac \pi2\right) \setminus \{1\}$$

My Approach:

$${\frac {d}{dx}}(e^{xlnlnx})+{\frac {d}{dx}}(e^{sinxln(arcsinx)})$$ $$= e^{xlnlnx}.{\frac {d}{dx}}(xlnlnx)+e^{sinxln(arcsinx)}.{\frac {d}{dx}}(sinxln(arcsinx))$$ $$=e^{xlnlnx}.(1.lnlnx+x{\frac {1}{lnx}})+e^{sinxln(arcsinx)}.cosx.ln(arcsinx)+sinx.{\frac {1}{arcsinx}}$$

There is something wrong I know, but I am not an experts and that's why posted here. Any help would be appreciated.

Second approach:

First part: $$y= lnx^x$$ $$lny = ln(lnx)^x$$ $${\frac {d}{dx}}lny = {\frac {d}{dx}}[ln(lnx)^x]$$ $${\frac 1y}{\frac {dy}{dx}} = {\frac {d}{dx}}[xln(lnx)]$$ $${\frac {dy}{dx}} = [ln(lnx)+{\frac {1}{lnx}}]lnx^x$$

Second part: $$y = (sin^{-1}x)^{sinx}$$ $$lny = ln(sin^{-1}x)^{sinx}$$ $${\frac {d}{dx}}lny = {\frac {d}{dx}}[sinx.ln(sin^{-}x)]$$ $${\frac 1y}{\frac {dy}{dx}} = {\frac {d}{dx}}sinx.ln(sin^{-1}x)+sinx.{\frac {d}{dx}}[ln(sin^{-1}x)]$$ $${\frac {dy}{dx}} = [cosx.ln(sin^1x)+{\frac {sinx}{sin^-1x(\sqrt {1-x^2})}}].sin^{-1}x^sinx$$

Running into an endless loop... any help?

  • 2
    Use log differentiation to find the derivatives of $(\ln)^x$ and $(\sin^{-1})^{\sin x}$ respectively, then add them together. – 2017-02-15
  • 1
    Please read my edit, to check if it is what you intended to post. Note: $]a, b[ = (a, b)$. – 2017-02-15

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Hint: you can transform it to exponent form $$y={ e }^{ x\ln { \ln { x } } }+{ e }^{ \sin { x } \ln { \left( \arcsin { x } \right) } }$$

  • 0
    tried your approach. any suggestion on next step? – 2017-02-16
  • 0
    Have you tried the chain rule? – 2017-02-16