174. Find the derivative of the function $y=x^{2}-3 x+5$.
Solution. $1^{0} . y_{\text {n }}=(x+\Delta x)^{2}-3(x+\Delta x)+5=x^{2}+2 x \Delta x+(\Delta x)^{2}-$ $-3 x-3 \Delta x+5$

$\left.2^{0} . \Delta y=y_{\mathrm{r}}-y=\left(x^{2}+2 x \Delta x\right)+(\Delta x)^{2}-3 x+3 \Delta x+5\right)-\left(x^{2}-3 x+5\right)=$ $=2 x \Delta x+(\Delta x)^{2}-3 \Delta x$.

$3^{0} . \frac{\Delta y}{\Delta x}=\frac{2 x \Delta x+(\Delta x)^{2}-3 \Delta x}{\Delta x}=2 x+\Delta x-3$.