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derivative of 5x^cos (x^x)
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****5x^cos(x^2)
Take the logarithm of both sides first, and then apply the derivative operator:\[y=5x^{\cos x^2} \rightarrow \ln y=\ln 5+\cos (x^2) \ln x\]
Then\[\frac{y'}{y}=\frac{\cos(x^2)}{x}+\ln x \times -\sin(x^2) \times 2x\]
Multiply both sides by y = 5x^(cos(x^2)) to get\[y'=5x^{\cos{x^2}}\left[ \frac{\cos (x^2)}{x}-2x \sin (x^2) \ln x \right]\]
Just check my working...rushing...
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