Partial Derivatives

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AP Calculus BC › Partial Derivatives

Questions 1 - 10

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

${-4\cdot 4^{(3z)}xycos(x^{2})}$

${8\cdot 4^{(6z)}x^{2}y^{3}cos(x^{2})^{2}}$

${- 2\cdot 4^{(3z)}ysin(x^{2}) - 2\cdot 4^{(3z)}xy^{2}cos(x^{2})}$

${4\cdot 4^{(6z)}xy^{3}cos(x^{2})sin(x^{2})}$

Explanation

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

Screen shot 2015 07 21 at 10.09.47 am

Given the above graph of , what is $\lim_{x\rightarrow 0^{-}}f(x)$ ?

$-\infty$

$\infty$

Explanation

Examining the graph, we want to find where the graph tends to as it approaches zero from the left hand side. We can see that there appears to be a vertical asymptote at zero. As the x values approach zero from the left, the function values of the graph tend towards negative infinity.

Therefore, we can observe that $\lim_{x\rightarrow 0^{-}}f(x)=-\infty$ as approaches from the left.

Evaluate the following limit:

$\lim_{x\rightarrow 2^{-}}f(x), f(x)=\left\{\begin{matrix} \ln(2-x), x\leq 2\ x-3, x> 2 \end{matrix}\right.$

$-\infty$

$\infty$

Explanation

To evaluate the limit, we must determine whether it is right or left sided. The negative sign "exponent" on the 2 indicates that we are approaching from the left, or numbers slightly less than 2. So, we choose the part of the piecewise function that is for x values less than or equal to 2. Now, evaluating the limit, as natural log approaches zero, we get $-\infty$ .

Find of the following function:

$f(x,y,z)=y^2\sec(z)-\cos(xy)$

$2y\sec(z)+x\sin(xy)$

$2y\sec(z)-x\sin(xy)$

$2y+x\sin(xy)$

Explanation

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivative of the function with respect to y is

$f_y=2y\sec(z)+x\sin(xy)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow1-}\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}\end{align*}$

$\infty$

$-\infty$

$\frac{41}{135424}$

$-\frac{41}{135424}$

Explanation

$\begin{align*}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}\rightarrow\frac{0}{0}\&\frac{82x - 82}{\frac{368}{x}}\rightarrow\frac{0}{368}=0\&lim_{x\rightarrow1-}\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}=0\end{align*}$

Screen shot 2015 07 21 at 10.09.47 am

Given the above graph of , what is $\lim_{x\rightarrow 0^{-}}f(x)$ ?

$-\infty$

$\infty$

Explanation

Therefore, we can observe that $\lim_{x\rightarrow 0^{-}}f(x)=-\infty$ as approaches from the left.

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(e^{(x)}ln(z))}{y}- 5e^{(x^{2})}e^{(y)}cos(z)\&\text{In the direction of }\overrightarrow{v}=(-4,-15,0).\end{align*}$

${4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)}$

${75e^{(x^{2})}e^{(y)}cos(z) -\frac{ (4e^{(x)}ln(z))}{y}+\frac{ (15e^{(x)}ln(z))}{y^{2}}+ 40xe^{(x^{2})}e^{(y)}cos(z)}$

${155xe^{(x^{2})}e^{(y)}sin(z) -\frac{ (15.5e^{(x)})}{(y^{2}z)}}$

${77.6e^{(x^{2})}e^{(y)}sin(z) - 77.6e^{(x^{2})}e^{(y)}cos(z) +\frac{ (15.5e^{(x)})}{(yz)}+\frac{ (15.5e^{(x)}ln(z))}{y}-\frac{ (15.5e^{(x)}ln(z))}{y^{2}}- 155xe^{(x^{2})}e^{(y)}cos(z)}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-15)^2+(0)^2}=15.52\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{15.52}=-0.26\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{15.52}=-0.97\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{15.52}=0\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.26)(\frac{(e^{(x)}ln(z))}{y}- 10xe^{(x^{2})}e^{(y)}cos(z))+(-0.97)(- 5e^{(x^{2})}e^{(y)}cos(z) -\frac{ (e^{(x)}ln(z))}{y^{2}})+(0)(5e^{(x^{2})}e^{(y)}sin(z) +\frac{ e^{(x)}}{(yz)})\&D_{\overrightarrow{u}}(x,y,z)=4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)\end{align*}$

$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow5-}\frac{(9x - 45)}{(17x - 85)}\end{align*}$

$\frac{9}{17}$

$\infty$

$-\infty$

Explanation

$\begin{align*}&\text{We cannot simply plug in the value of }x=5\text{, as that would}\&\text{create a zero value in the denominator, and not tell us much as is.}\&\text{However, note that the expression}\&\frac{(9x - 45)}{(17x - 85)}\&\text{can be factored. This will allow us to find the limit}\&\text{taken from the left:}\&\frac{(9\cdot (x - 5))}{(17\cdot (x - 5))}\&\frac{9}{17}\&lim_{x\rightarrow5-}\frac{(9x - 45)}{(17x - 85)}=\frac{9}{17}\text{Another method, since both numerator and denominator go}\&\text{to zero initially, is to use L'Hopital's rule:}\&lim_{x\rightarrow5-}\frac{(9)}{(17)}=\frac{9}{17}\end{align*}$

Find the partial derivative of the function $f(x,y,z)=ze^{x^2+y}$ .

$f_x=2xze^{x^2+y}$

$f_x=yze^{x^2+y}$

$f_x=e^{x^2+y}$

$f_x=ze^{2x}$

Explanation

To find the partial derivative of the function $f(x,y,z)=ze^{x^2+y}$ , we take the derivative with respect to $\small x$ while holding $\small y,z$ constant. We also use the chain rule to get

$\small \small \small f_x=\frac{\partial}{\partial x}f(x,y,z)=\frac{\partial}{\partial x}ze^{x^2+y}=ze^{x^2+y}\frac{\partial}{\partial x}(x^2+y)$

$\small =2xze^{x^2+y}$

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

${-\frac{(12ze^{(3x)}e^{(3y)})}{7}}$

${\frac{(48z^{2}e^{(6x)}e^{(6y)})}{49}}$

${-\frac{ (4ze^{(3x)}e^{(3y)})}{7}-\frac{ (6z^{2}e^{(3x)}e^{(3y)})}{7}}$

${\frac{(24z^{3}e^{(6x)}e^{(6y)})}{49}}$

Explanation

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

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