AP Calculus BC - Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\&\text{Where }f(x,y,z)=\frac{(y^{2}tan(11x + x^{2}))}{(3z)}- 2cos(z^{2})sin(x^{3})e^{(y^{3})}\end{align*}$

${\frac{(2y\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 18x^{2}y^{2}cos(x^{3})cos(z^{2})e^{(y^{3})}}$

${(\frac{(2y\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 18x^{2}y^{2}cos(x^{3})cos(z^{2})e^{(y^{3})})\cdot (\frac{(y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})})}$

${\frac{(2ytan(11x + x^{2}))}{(3z)}+\frac{ (y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})} - 6y^{2}cos(z^{2})sin(x^{3})e^{(y^{3})}}$

${(\frac{(2ytan(11x + x^{2}))}{(3z)}- 6y^{2}cos(z^{2})sin(x^{3})e^{(y^{3})})\cdot (\frac{(y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})})}$

Explanation

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[sin(u)]=cos(u)du\&d[e^u]=e^udu\&\text{Solve step-by-step:}\&f(x,y,z)=\frac{(y^{2}tan(11x + x^{2}))}{(3z)}- 2cos(z^{2})sin(x^{3})e^{(y^{3})}\&f_{x}=\frac{(y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})}\&f_{xy}=\frac{(2y\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 18x^{2}y^{2}cos(x^{3})cos(z^{2})e^{(y^{3})}\end{align*}$

Evaluate the limit

$\lim_{n \to 1}\frac{n^2-2n+1}{n^2-1}$

$\infty$

$-\infty$

Explanation

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=1; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\lim_{n \to 1}\frac{n^2-2n+1}{n^2-1}=\lim_{n \to 1}\frac{(n-1)(n-1)}{(n-1)(n+1)}=\lim_{n \to 1}\frac{n-1}{n+1}=\frac{1-1}{1+1}=\frac{0}{2}=0$

Find the total derivative of the following function:

$f(x, y,z)=x^2y+\csc(z)$

$2xydx+x^2dy-\csc(z)\cot(z)dz$

$2xydx+x^2dy+\csc(z)\cot(z)dz$

$2xdx-\csc(z)\cot(z)dz$

$2xy+x^2-\csc(z)\cot(z)$

Explanation

The total derivative of a function is given by

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

The partial derivatives are

$f_z=-\csc(z)\cot(z)$

The derivatives were found using the following rules:

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

Evaluate the limit:

$\lim_{n \to \infty}\frac{n^3}{4n^3+n^2+n}$

$\frac{1}{4}$

$\infty$

$-\infty$

Explanation

To evaluate the limit, we must factor out a term consisting of the highest power term divided by itself (which equals one, so we aren't changing the original function):

$\lim_{n \to \infty}(\frac{n^3}{n^3})\frac{1}{4+n^{-1}+n^{-2}}$

The term we factored goes to one, and the two terms with negative exponents in the denominator go to zero (they are each "fractions" with n in their denominator - the terms go to zero as the denominator goes to infinity), so we are left with $\frac{1}{4}$ .

Find $f_{xy}$ of the function $f(x,y,z)=3y\sin(x)+8x^2yz^3$

$3\cos(x)+16xz^3$

$5\cos(x)+16x^2z^4$

$3\cos(x)+14xz^3$

$3\sin(x)+16xz^3$

Explanation

To find $f_{xy}$ of the function, you take two consecutive partial derivatives:

$\frac{\partial }{\partial x}(3y\sin(x)+8x^2yz^3)=3y\cos(x)+16xyz^3$

$\frac{\partial }{\partial y}(3y\cos(x)+16xyz^3)=3\cos(x)+16xz^3$

Evaluate the limit

$\lim_{n \to 1}\frac{n^2-2n+1}{n^2-1}$

$\infty$

$-\infty$

Explanation

$\lim_{n \to 1}\frac{n^2-2n+1}{n^2-1}=\lim_{n \to 1}\frac{(n-1)(n-1)}{(n-1)(n+1)}=\lim_{n \to 1}\frac{n-1}{n+1}=\frac{1-1}{1+1}=\frac{0}{2}=0$

$\begin{align*}&\text{Calculate the limits of the function}\&\lim_{(x,y)\rightarrow(0+,0+)}\frac{(17x^{2}y^{2} + 14xy^{3} - 5y^{4})}{(8x^{3}y - 15x^{2}y^{2} + 3x^{4} + 2y^{4})}\&\text{along the y-axis and the line, }y=x\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{5}{2}\&y=x:-13\end{align*}$

$\begin{align*}&\text{y-axis: }-156\&y=x:-\frac{95}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }117\&y=x:45\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{5}{34}\&y=x:-\frac{13}{4}\end{align*}$

Explanation

$\begin{align*}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&\frac{(17(0)^{2}y^{2} + 14(0)y^{3} - 5y^{4})}{(8(0)^{3}y - 15(0)^{2}y^{2} + 3(0)^{4} + 2y^{4})}=\frac{-5y^{4}}{2y^{4}}\&lim_{(0,y)\rightarrow(0+,0+)}\frac{-5y^{4}}{2y^{4}}=-\frac{5}{2}\&\text{We can make a similar substitution for }y=x:\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(17x^{2}(x)^{2} + 14x(x)^{3} - 5(x)^{4})}{(8x^{3}(x) - 15x^{2}(x)^{2} + 3x^{4} + 2(x)^{4})}=-13\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{3}cos(z^{4})e^{(x^{4})}\&\text{In the direction of }\overrightarrow{v}=(0,-15,-12).\end{align*}$

${2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}}$

${48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 45y^{2}cos(z^{4})e^{(x^{4})}}$

${-922x^{3}y^{2}z^{3}sin(z^{4})e^{(x^{4})}}$

${57.6y^{2}cos(z^{4})e^{(x^{4})} + 76.8x^{3}y^{3}cos(z^{4})e^{(x^{4})} - 76.8y^{3}z^{3}sin(z^{4})e^{(x^{4})}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-15)^2+(-12)^2}=19.21\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.21}=0\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{19.21}=-0.78\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{19.21}=-0.62\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}y^{3}cos(z^{4})e^{(x^{4})})+(-0.78)(3y^{2}cos(z^{4})e^{(x^{4})})+(-0.62)(-4y^{3}z^{3}sin(z^{4})e^{(x^{4})})\&D_{\overrightarrow{u}}(x,y,z)=2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}\end{align*}$

What is the partial derivative of the function

$f(x,y)=e^{xy}$ ?

$f_x(x,y)=ye^{xy}$

$f_x(x,y)=xye^{xy}$

$f_x(x,y)=xe^{xy}$

$f_x(x,y)=e^{xy}$

Explanation

We can find given $f(x,y)=e^{xy}$ by differentiating the function while holding constant, i.e. we treat as a number. So we get

$f_x(x,y)=\frac{\partial}{\partial x}e^{xy}=e^{xy}\frac{\partial}{\partial x}xy=e^{xy}\cdot y=ye^{xy}$

Evaluate the following limit:

$\lim_{x\rightarrow \infty}\frac{x^4}{x^3+5x^2+x}$

$\infty$

$-\infty$

$\frac{1}{5}$

Explanation

To evaluate the limit, we must first pull out a factor consisting of the highest power term divided by itself (one, essentially):

$\lim_{x\rightarrow \infty} \frac{x^4}{x^4}(\frac{1}{x^{-1}+5x^{-2}+x^{-3}})$

After the factor we created becomes 1, the negative exponent terms go to zero as x approaches infinity, therefore we are left with $\frac{1}{0}=\infty$ .

Partial Derivatives

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