y^2=x+√x
両辺xで微分して、
2yy'=1+1/(2√x)=(1+2√x)/(2√x)
y'=(1+2√x)/(4y√x)
=(1+2√x)/{4√(x^2+x√x)}
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y^2=x+√x
両辺xで微分して、
2yy'=1+1/(2√x)=(1+2√x)/(2√x)
y'=(1+2√x)/(4y√x)
=(1+2√x)/{4√(x^2+x√x)}
Users viewing this question
are also looking at these questions 😉