Let f:R−0→R be a function satisfying f(yx)=f(y)f(x) for all x,y,f(y)=0. If f′(1)=2024, then
Solutions · 1
Official solution
\[f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}\]
\[f′(1)=2024f(1)=1\]
Partially differentiating w. r. t. x
\[f′(yx)⋅y1=f(y)1f′(x)y→xf′(1)⋅x1=f(x)f′(x)2024f(x)=xf′(x)⇒xf′(x)−2024f(x)=0\]