Question

find the indicated partial derivatives of (f(x,y)=7x^{3}y^{2}) using the limit definition. the limits need to be reduced as much as possible before they are evaluated.
(f_{x}(x,y)=lim_{h
ightarrow0}\frac{f(x + h,y)-f(x,y)}{h}=lim_{h
ightarrow0}(quad)=quad
(f_{y}(x,y)=lim_{h
ightarrow0}\frac{f(x,y + h)-f(x,y)}{h}=lim_{h
ightarrow0}(quad)=quad
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Explanation:

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

Given \(f(x,y)=7x^{3}y^{2}\), then \(f(x + h,y)=7(x + h)^{3}y^{2}\). Expand \((x + h)^{3}\) using the formula \((a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3}\), so \(f(x + h,y)=7(x^{3}+3x^{2}h + 3xh^{2}+h^{3})y^{2}=7x^{3}y^{2}+21x^{2}hy^{2}+21xh^{2}y^{2}+7h^{3}y^{2}\).

Step2: Calculate \(f_x(x,y)\) using the limit - definition

\[

$$\begin{align*} f_x(x,y)&=\lim_{h ightarrow0}\frac{f(x + h,y)-f(x,y)}{h}\\ &=\lim_{h ightarrow0}\frac{(7x^{3}y^{2}+21x^{2}hy^{2}+21xh^{2}y^{2}+7h^{3}y^{2})-7x^{3}y^{2}}{h}\\ &=\lim_{h ightarrow0}\frac{21x^{2}hy^{2}+21xh^{2}y^{2}+7h^{3}y^{2}}{h}\\ &=\lim_{h ightarrow0}(21x^{2}y^{2}+21xh y^{2}+7h^{2}y^{2}) \end{align*}$$

\]
As \(h
ightarrow0\), \(f_x(x,y)=21x^{2}y^{2}\).

Step3: Find \(f(x,y + h)\)

\(f(x,y + h)=7x^{3}(y + h)^{2}\). Expand \((y + h)^{2}\) using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), so \(f(x,y + h)=7x^{3}(y^{2}+2yh+h^{2})=7x^{3}y^{2}+14x^{3}yh + 7x^{3}h^{2}\).

Step4: Calculate \(f_y(x,y)\) using the limit - definition

\[

$$\begin{align*} f_y(x,y)&=\lim_{h ightarrow0}\frac{f(x,y + h)-f(x,y)}{h}\\ &=\lim_{h ightarrow0}\frac{(7x^{3}y^{2}+14x^{3}yh + 7x^{3}h^{2})-7x^{3}y^{2}}{h}\\ &=\lim_{h ightarrow0}\frac{14x^{3}yh + 7x^{3}h^{2}}{h}\\ &=\lim_{h ightarrow0}(14x^{3}y+7x^{3}h) \end{align*}$$

\]
As \(h
ightarrow0\), \(f_y(x,y)=14x^{3}y\).

Answer:

\(f_x(x,y)=21x^{2}y^{2}\), \(f_y(x,y)=14x^{3}y\)