Partial Derivatives - AP Calculus BC

Card 0 of 7240

Question

$\begin{align}&\text{Perform the partial derivation, }f_{yx}\&\text{Where }f(x,y,z)=-\frac{(2\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{2})}\end{align}$

Answer

$\begin{align}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&\text{Solve step-by-step:}\&f(x,y,z)=-\frac{(2\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{2})}\&f_{y}=\frac{(4\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{3})}\&f_{yx}=\frac{(12\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{3})}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow0-}\frac{x^{2}}{sin(13\cdot \pi x)^{2}}\end{align}$

Answer

$\begin{align}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{x^{2}}{sin(13\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\&\frac{2x}{26\cdot \pi cos(13\cdot \pi x)sin(13\cdot \pi x)}\rightarrow\frac{0}{0}\&\frac{2}{338\cdot \pi ^{2}cos(13\cdot \pi x)^{2} - 338\cdot \pi ^{2}sin(13\cdot \pi x)^{2}}\rightarrow\frac{2}{338\cdot \pi ^{2}}\&lim_{x\rightarrow0-}\frac{x^{2}}{sin(13\cdot \pi x)^{2}}=\frac{1}{(169\cdot \pi ^{2})}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(32x^{2}y + 10y^{3})}{(6x^{3} - 15x^{2}y + 13y^{3})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(32(0)^{2}y + 10y^{3})}{(6(0)^{3} - 15(0)^{2}y + 13y^{3})}=\frac{-10y^{3}}{13y^{3}}\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{-10y^{3}}{13y^{3}}=-\frac{10}{13}\&\text{We can make a similar substitution for }y=x:\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(32x^{2}(x) + 10(x)^{3})}{(6x^{3} - 15x^{2}(x) + 13(x)^{3})}=-\frac{21}{2}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(18x^{3}y^{3})}{(x^{5}y - 14xy^{5} - 17x^{3}y^{3} + 4x^{6} + 5y^{6})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(18(0)^{3}y^{3})}{((0)^{5}y - 14(0)y^{5} - 17(0)^{3}y^{3} + 4(0)^{6} + 5y^{6})}=\frac{0}{5y^{6}}\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{5y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(18x^{3}(x)^{3})}{(x^{5}(x) - 14x(x)^{5} - 17x^{3}(x)^{3} + 4x^{6} + 5(x)^{6})}=\frac{6}{7}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Perform the partial derivation, }f_{yy}\&\text{Where }f(x,y,z)=2sin(y^{2})ln(x)sin(z) - 4e^{(3y)}cos(x^{2} - 3x)cos(z)\end{align}$

Answer

$\begin{align}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[sin(u)]=cos(u)du\&d[cos(u)]=-sin(u)du\&\text{Solve step-by-step:}\&f(x,y,z)=2sin(y^{2})ln(x)sin(z) - 4e^{(3y)}cos(x^{2} - 3x)cos(z)\&f_{y}=4ycos(y^{2})ln(x)sin(z) - 12e^{(3y)}cos(x^{2} - 3x)cos(z)\&f_{yy}=4cos(y^{2})ln(x)sin(z) - 36e^{(3y)}cos(x^{2} - 3x)cos(z) - 8y^{2}sin(y^{2})ln(x)sin(z)\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0+,0+)}-\frac{(8xy^{4} - 15x^{2}y^{3} + 7x^{5})}{(5x^{5} + y^{5})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(8(0)y^{4} - 15(0)^{2}y^{3} + 7(0)^{5})}{(5(0)^{5} + y^{5})}=\frac{0}{y^{5}}\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{5}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(8x(x)^{4} - 15x^{2}(x)^{3} + 7x^{5})}{(5x^{5} + (x)^{5})}=0\&\text{Since the two limits are equal, it's possible the limit exists.}\&\text{Testing additional lines could prove or disprove this.}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(15x^{5}y - 20x^{3}y^{3} + 10x^{6})}{(2x^{6} - x^{4}y^{2} + y^{6})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&\frac{(15(0)^{5}y - 20(0)^{3}y^{3} + 10(0)^{6})}{(2(0)^{6} - (0)^{4}y^{2} + y^{6})}=\frac{0}{y^{6}}\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(15x^{5}(x) - 20x^{3}(x)^{3} + 10x^{6})}{(2x^{6} - x^{4}(x)^{2} + (x)^{6})}=\frac{5}{2}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Perform the partial derivation, }f_{xz}\&\text{Where }f(x,y,z)=y^{2}sin(x^{3})e^{(z)} +\frac{ (2^zx^{2}e^{(y)})}{4}\end{align}$

Answer

$\begin{align}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[a^u]=a^uduln(a)\&d[sin(u)]=cos(u)du\&d[e^u]=e^udu\&\text{Solve step-by-step:}\&f(x,y,z)=y^{2}sin(x^{3})e^{(z)} +\frac{ (2^zx^{2}e^{(y)})}{4}\&f_{x}=\frac{(2^zxe^{(y)})}{2}+ 3x^{2}y^{2}cos(x^{3})e^{(z)}\&f_{xz}=3x^{2}y^{2}cos(x^{3})e^{(z)} +\frac{ (2^zxe^{(y)}ln(2))}{2}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(xy^{3} + x^{3}y - 2x^{4})}{(x^{4} + y^{4})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&-\frac{((0)y^{3} + (0)^{3}y - 2(0)^{4})}{((0)^{4} + y^{4})}=\frac{0}{y^{4}}\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{y^{4}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(x(x)^{3} + x^{3}(x) - 2x^{4})}{(x^{4} + (x)^{4})}=0\&\text{Since the two limits are equal, it's possible the limit exists.}\&\text{Testing additional lines could prove or disprove this.}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(14y^{4})}{(10x^{2}y^{2} + 8x^{3}y + 9x^{4} + 2y^{4})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(14y^{4})}{(10(0)^{2}y^{2} + 8(0)^{3}y + 9(0)^{4} + 2y^{4})}=\frac{-14y^{4}}{2y^{4}}\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{-14y^{4}}{2y^{4}}=-7\&\text{We can make a similar substitution for }y=x:\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(14(x)^{4})}{(10x^{2}(x)^{2} + 8x^{3}(x) + 9x^{4} + 2(x)^{4})}=-\frac{14}{29}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Perform the partial derivation, }f_{yy}\&\text{Where }f(x,y,z)=-\frac{ 4}{(y^{3}z^{2})}-\frac{ (z^{2}e^{(y^{2})}sin(x))}{3}\end{align}$

Answer

$\begin{align}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&\text{Solve step-by-step:}\&f(x,y,z)=-\frac{ 4}{(y^{3}z^{2})}-\frac{ (z^{2}e^{(y^{2})}sin(x))}{3}\&f_{y}=\frac{12}{(y^{4}z^{2})}-\frac{ (2yz^{2}e^{(y^{2})}sin(x))}{3}\&f_{yy}=-\frac{ 48}{(y^{5}z^{2})}-\frac{ (2z^{2}e^{(y^{2})}sin(x))}{3}-\frac{ (4y^{2}z^{2}e^{(y^{2})}sin(x))}{3}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(18xy^{5} + 4x^{5}y)}{(11x^{3}y^{3} + xy^{5} - x^{6} - 4y^{6})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(18(0)y^{5} + 4(0)^{5}y)}{(11(0)^{3}y^{3} + (0)y^{5} - (0)^{6} - 4y^{6})}=\frac{0}{4y^{6}}\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{4y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(18x(x)^{5} + 4x^{5}(x))}{(11x^{3}(x)^{3} + x(x)^{5} - x^{6} - 4(x)^{6})}=-\frac{22}{7}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

Compute $\frac{df}{dt}$ for the function $f(x,y,z)=3xe^{z+y}$ where the variables , and are functions of the of the parameter :

$x = t: : :: : : :y=\ln(t): : : : : : : :z=t^2$

Answer

Compute $\frac{df}{dt}$ for the function $f(x,y,z)=3xe^{z+y}$ where the variables , and are functions of the of the parameter :

$x = t: : :: : : :y=\ln(t): : : : : : : :z=t^2$

Because each function is given in terms of the parameter we can conveniently write the function as a function of alone:

Solution 1

For the function described in this problem, the function can be written strictly in terms of the parameter by simply substituting the definitions given for , or .

$f(x,y,z)=f(t, : \ln t,: :t^2 ) )=3te^{t^2+\ln t}=3t^2e^{t^2}$

$f(t)=3t^2e^{t^2}$

So now we have the function written in the form of a single variable function of We can use the usual chain-rule.

Apply the product rule and and the chain-rule:

$\frac{df}{dt}=3t^2\left(\frac{d}{dt}e^{t^2} \right )+e^{t^2}\left(\frac{d}{dt}3t^2 \right )=6te^{t^2}+3t^2e^{t^2}(2t)$

$=6te^{t^2}+6t^3e^{t^2}$

$=6te^{t^2}(1+t^2)$

Solution 2

The derivative of with respect to will be the sum of the derivatives of each variable , , and with respect to

$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}$ (1)

In Equation (1) we use the partial derivative notation for a derivative of with respect to the variables , , and since it is a multi-variable function, we have to specify which variable we are differentiating unless we are differentiating with respect to

We return to the standard derivative notation when computing the derivatives with respect to since each function , , , and can be written as a single variable function of Now simply apply (3) to the function term-by-term:

$\frac{\partial f}{\partial x}\frac{dx}{dt}=3e^{z+y}$

$\frac{\partial f}{\partial y}\frac{dy}{dt}=3xe^{z+y}\frac{1}{t}$

$\frac{\partial f}{\partial z}\frac{dz}{dt}=3xe^{z+y}(2t)=6xte^{z+y}$

Now ad the terms to get an expression for $\frac{df}{dt}$ and then rewrite in terms of

$\frac{df}{dt}=3xe^{z+y}\frac{1}{t}+6xte^{z+y}+3e^{z+y}$

$=3e^{z+y}(1+\frac{x}{t}+2xt)$

$=3e^{t^2+\ln t}\left(1+\frac{t}{t}+2t^2\right)$

$=6e^{t^2+\ln t}\left(1+t^2\right)$

$=6te^{t^2}\left(1+t^2\right)$

Therefore:

$\frac{df}{dt}=6te^{t^2}\left(1+t^2\right)$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0-,0+)}\frac{(11x^{4}y^{2} - 17x^{6})}{(2x^{4}y^{2} - 12x^{3}y^{3} - 19x^{5}y + 17x^{6} + y^{6})}\&\text{along the x-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\&\frac{(11(0)^{4}y^{2} - 17(0)^{6})}{(2(0)^{4}y^{2} - 12(0)^{3}y^{3} - 19(0)^{5}y + 17(0)^{6} + y^{6})}=\frac{0}{y^{6}}\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(11x^{4}(x)^{2} - 17x^{6})}{(2x^{4}(x)^{2} - 12x^{3}(x)^{3} - 19x^{5}(x) + 17x^{6} + (x)^{6})}=\frac{6}{11}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Perform the partial derivation, }f_{yz}\&\text{Where }f(x,y,z)=- 3ln(z^{2})e^{(x^{2})} -\frac{ (2\cdot 3^{(3z)}\cdot 3^ysin(x^{3} - 3x))}{7}\end{align}$

Answer

$\begin{align}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[a^u]=a^uduln(a)\&d[ln(u)]=\frac{du}{u}\&\text{Solve step-by-step:}\&f(x,y,z)=- 3ln(z^{2})e^{(x^{2})} -\frac{ (2\cdot 3^{(3z)}\cdot 3^ysin(x^{3} - 3x))}{7}\&f_{y}=-\frac{(2\cdot 3^{(3z)}\cdot 3^yln(3)sin(x^{3} - 3x))}{7}\&f_{yz}=-\frac{(6\cdot 3^{(3z)}\cdot 3^yln(3)^{2}sin(x^{3} - 3x))}{7}\end{align}$

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Question

$\begin{align}&\text{Perform the partial derivation, }f_{yz}\&\text{Where }f(x,y,z)=2cos(x^{2})e^{(z)}sin(3y + y^{3}) -\frac{ (3ln(y^{2}))}{z^{3}}\end{align}$

Answer

$\begin{align}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\&d[e^u]=e^udu\&d[ln(u)]=\frac{du}{u}\&\text{Solve step-by-step:}\&f(x,y,z)=2cos(x^{2})e^{(z)}sin(3y + y^{3}) -\frac{ (3ln(y^{2}))}{z^{3}}\&f_{y}=2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3) -\frac{ 6}{(yz^{3})}\&f_{yz}=\frac{18}{(yz^{4})}+ 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3)\end{align}$

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Question

$\begin{align}&\text{Perform the partial derivation, }f_{xx}\&\text{Where }f(x,y,z)=\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (4ln(x^{2})sin(y))}{z^{2}}\end{align}$

Answer

$\begin{align}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[ln(u)]=\frac{du}{u}\&\text{Solve step-by-step:}\&f(x,y,z)=\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (4ln(x^{2})sin(y))}{z^{2}}\&f_{x}=\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (8sin(y))}{(xz^{2})}\&f_{xx}=\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}+\frac{ (8sin(y))}{(x^{2}z^{2})}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(9x^{4}y^{2} + 9xy^{5} - 18x^{6})}{(3x^{6} + 2y^{6})}\&\text{along the x-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(9(0)^{4}y^{2} + 9(0)y^{5} - 18(0)^{6})}{(3(0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{2y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(9x^{4}(x)^{2} + 9x(x)^{5} - 18x^{6})}{(3x^{6} + 2(x)^{6})}=0\&\text{Since the two limits are equal, it's possible the limit exists.}\&\text{Testing additional lines could prove or disprove this.}\end{align}$

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Question

Calculate the partial derivative with respect to y, fy(x, y), of the following function:

$f(x, y)=xy+3x^2y+\sin(xy)$

Answer

The partial derivative of f(x,y) is calculated by treating x as a constant. The third term requires an application of the chain rule:

$\frac{\partial f(x,y)}{\partial y} =x+3x^2+\cos(xy)x=x[1+3x+\cos(xy)]$

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Question

Calculate the partial derivative with respect to of the following function:

$f(x,y,z)=7e^x^z\ln(y)$

Answer

When calculating the partial derivative with respect to the variable of a function of more than one variable, apply the standard rules for differentiating a function of a single variable, and treat the other variables as constants. In this case, we have:

$f(x,y,z)=7e^x^z\ln(y)$

We are being asked to differentiate with respect to , so we treat the variables and as constants, recognize that the term is now just a constant, and apply the rule of differentiation for the natural logarithm to find the partial derivative $\frac{\partial f}{\partial y}$ , as shown:

$\frac{\partial f}{\partial y}=7e^x^z(\ln(y))'$

$=7e^x^z(\frac{1}{y})$

$=\frac{7e^x^z}{y}$

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