7.4.39-eg - ( x )) = ln x ( x +1) , and then simplifying...

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Example Find y 0 ( x ) if y ( x ) = x ( x +1) . Two (related) solutions: 1. Note that we can rewrite y equivalently as y ( x ) = e ln x ( x +1) = e ( x +1) ln x . Now differentiate using chain rule where f ( x ) = e x and g ( x ) = ( x + 1) ln x . Since f 0 ( x ) = e x and g 0 ( x ) = ln x + x + 1 x , dy dx = e ( x +1) ln x · ± ln x + x + 1 x ² = x ( x +1) · ± ln x + 1 + 1 x ² . 2. We can also use logarithmic differentiation as a way to remove the difficult exponent in this function. That is, apply the natural log function to both sides of this equation: y ( x ) = x ( x +1) , to obtain ln( y
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Unformatted text preview: ( x )) = ln x ( x +1) , and then simplifying the right-hand side, ln( y ( x )) = ( x + 1) ln x. Now dierentiate this last equation implicitly: 1 y ( x ) y ( x ) = ( x + 1) 1 x + 1 ln x, so y ( x ) = y ( x ) x + 1 x + ln x . Now substituting back in the value of y ( x ) = x ( x +1) , we have y ( x ) = x ( x +1) ln x + 1 + 1 x ....
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