Question

for the function $f(x,y)=4x^{2}+3y^{2}$, find $\frac{f(x + h,y)-f(x,y)}{h}$.
$\frac{f(x + h,y)-f(x,y)}{h}=\square$

Explanation:

Step1: Find \( f(x + h,y) \)

Substitute \( x\) with \(x + h\) in \(f(x,y)=4x^{2}+3y^{2}\).
\(f(x + h,y)=4(x + h)^{2}+3y^{2}=4(x^{2}+2xh+h^{2})+3y^{2}=4x^{2}+8xh + 4h^{2}+3y^{2}\)

Step2: Calculate \( f(x + h,y)-f(x,y) \)

\(f(x + h,y)-f(x,y)=(4x^{2}+8xh + 4h^{2}+3y^{2})-(4x^{2}+3y^{2})=8xh+4h^{2}\)

Step3: Divide by \(h\)

\(\frac{f(x + h,y)-f(x,y)}{h}=\frac{8xh + 4h^{2}}{h}\). Factor out \(h\) in the numerator: \(\frac{h(8x + 4h)}{h}\). Cancel out \(h\) (\(h
eq0\)).

Answer:

\(8x + 4h\)