Differentiate $y = {(sin (x))}^{log (x)}$ $y=(sin(x))^(log(x))$ ?

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Differentiate $y = {(sin (x))}^{log (x)}$ $y=(sin(x))^(log(x))$ ?

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Answer 1 · 2018-01-02T06:47:15

$\frac{d y}{d x} = 0.4343 {(sin x)}^{log x} (\frac{sin x}{x} + cot x ln x)$ $(dy)/(dx)=0.4343(sinx)^(logx)(sinx/x+cotxlnx)$

Explanation:

As $y = {(sin (x))}^{log (x)}$ $y=(sin(x))^(log(x))$

we have $ln y = {ln (sin (x))}^{log (x)} = log x ln (sin x) = \frac{ln x}{ln 10} ln (sin x)$ $lny=ln(sin(x))^(log(x))=logxln(sinx)=lnx/ln10ln(sinx)$

Now as $ln y = \frac{ln x}{ln 10} ln (sin x)$ $lny=lnx/ln10ln(sinx)$

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{ln 10} (\frac{1}{x} ln (sin x) + ln x \cdot \frac{1}{sin x} \cdot cos x)$ $1/y(dy)/(dx)=1/ln10(1/xln(sinx)+lnx*1/sinx*cosx)$

= $\frac{1}{2.3026} (\frac{sin x}{x} + cot x ln x)$ $1/2.3026(sinx/x+cotxlnx)$

or $\frac{d y}{d x} = \frac{1}{2.3026} (\frac{sin x}{x} + cot x ln x) \times {(sin x)}^{log x}$ $(dy)/(dx)=1/2.3026(sinx/x+cotxlnx)xx(sinx)^(logx)$

= $0.4343 {(sin x)}^{log x} (\frac{sin x}{x} + cot x ln x)$ $0.4343(sinx)^(logx)(sinx/x+cotxlnx)$

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Differentiate $y = {(sin (x))}^{log (x)}$ $y=(sin(x))^(log(x))$ ?

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Differentiate $y = {(sin (x))}^{log (x)}$ $y=(sin(x))^(log(x))$ ?