For the function $f(x, y)=16 x y^{2}$, find $\frac{f(x+h, y)-f(x, y)}{h}$.
\[
\frac{f(x+h, y)-f(x, y)}{h}=\square
\]
Answer
Simplify the expression by canceling out the common factor of $h$: $\frac{f(x+h, y)-f(x, y)}{h} = \frac{16hy^2}{h} = 16y^2$
Step 1 ：Substitute $f(x+h, y)$ into the numerator: $f(x+h, y) = 16(x+h)y^2$
Step 2 ：Substitute $f(x, y)$ into the numerator: $f(x, y) = 16xy^2$
Step 3 ：Substitute the values into the expression: $\frac{f(x+h, y)-f(x, y)}{h} = \frac{16(x+h)y^2 - 16xy^2}{h}$
Step 4 ：Simplify the numerator: $16(x+h)y^2 - 16xy^2 = 16xy^2 + 16hy^2 - 16xy^2 = 16hy^2$
Step 5 ：Substitute the simplified numerator back into the expression: $\frac{f(x+h, y)-f(x, y)}{h} = \frac{16hy^2}{h}$
Step 6 ：Simplify the expression by canceling out the common factor of $h$: $\frac{f(x+h, y)-f(x, y)}{h} = \frac{16hy^2}{h} = 16y^2$
