Apply d d x {\displaystyle {\frac {d}{dx}}} to both sides of the equation x y = 5 y − x + 1 {\displaystyle x^{y}=5y-x+1} .
Differentiating the left-hand side requires work. We use logarithmic differentiation. Since log ( x y ) = y log x {\displaystyle \log(x^{y})=y\log x} , we get the derivative of x y {\displaystyle x^{y}} as
d d x ( x y ) = x y d d x log ( x y ) = x y d d x ( y log x ) = x y ( y ′ log x + y x ) . {\displaystyle {\frac {d}{dx}}(x^{y})=x^{y}{\frac {d}{dx}}\log(x^{y})=x^{y}{\frac {d}{dx}}(y\log x)=x^{y}\left(y'\log x+{\frac {y}{x}}\right).}
This implies that
x y ( y ′ log x + y x ) = d d x ( x y ) = d d x ( 5 y − x + 1 ) = 5 y ′ − 1. {\displaystyle x^{y}\left(y'\log x+{\frac {y}{x}}\right)={\frac {d}{dx}}(x^{y})={\frac {d}{dx}}(5y-x+1)=5y'-1.}
Now plugging x = 1 , y = 1 5 , {\displaystyle x=1,y={\frac {1}{5}},} we have 1 ⋅ ( y ′ ⋅ 0 + 1 5 1 ) = 5 y ′ − 1 {\displaystyle 1\cdot \left(y'\cdot 0+{\frac {\frac {1}{5}}{1}}\right)=5y'-1} .
So d y d x | x = 1 , y = 1 5 = y ′ | x = 1 , y = 1 5 = 6 25 {\displaystyle \left.{\frac {dy}{dx}}\right|_{x=1,y={\frac {1}{5}}}=\left.y'\right|_{x=1,y={\frac {1}{5}}}=\color {blue}{\frac {6}{25}}} .
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