Partial Derivatives

AP Calculus BC · Learn by Concept

Help Questions

AP Calculus BC › Partial Derivatives

1 - 10

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

${\frac{(12\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{3})}}$

CORRECT

${\frac{(48\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{6})}}$

${\frac{(4\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{3})}-\frac{ (6\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{2})}}$

${-\frac{(24\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{5})}}$

Explanation

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

Find the value of for at

CORRECT

Explanation

Note that for this problem, we're told to take the derivative with respect to one particular variable. This is known as taking a partial derivative; often it is denoted with the Greek character delta, $\delta$ , or by the subscript of the variable being considered such as or .

Taking the partial derivative of at

We find:

$f_x=6x^2y+4 \rightarrow72+4=76$

Evaluate the limit:

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

$\frac{1}{4}$

CORRECT

$\infty$

$-\infty$

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 $\frac{1}{4}$ .

$\begin{align*}&\text{Perform the partial derivation, }f_{xz}\&\text{Where }f(x,y,z)=\frac{{(2^{(4y)}\cdot2^{(3z)}cos{(4x)})}}{2}\end{align*}$

${-6\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}ln{(2)}}$

CORRECT

${12\cdot2^{(8y)}\cdot2^{(6z)}sin{(4x)}^2ln{(2)}}$

${\frac{{(3\cdot2^{(4y)}\cdot2^{(3z)}cos{(4x)}ln{(2)})}}{2} - 2\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}}$

${-3\cdot2^{(8y)}\cdot2^{(6z)}cos{(4x)}sin{(4x)}ln{(2)}}$

Explanation

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[a^u]=a^uduln(a)\&\text{Solve step-by-step:}\&f(x,y,z)=\frac{{(2^{(4y)}\cdot2^{(3z)}cos{(4x)})}}{2}\&f_{x}=-2\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}\&f_{xz}=-6\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}ln{(2)}\end{align*}$

$\begin{align*}&\text{Perform the partial derivation, }f_{xz}\&\text{Where }f(x,y,z)=\frac{{(2^{(4y)}\cdot2^{(3z)}cos{(4x)})}}{2}\end{align*}$

${-6\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}ln{(2)}}$

CORRECT

${12\cdot2^{(8y)}\cdot2^{(6z)}sin{(4x)}^2ln{(2)}}$

${\frac{{(3\cdot2^{(4y)}\cdot2^{(3z)}cos{(4x)}ln{(2)})}}{2} - 2\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}}$

${-3\cdot2^{(8y)}\cdot2^{(6z)}cos{(4x)}sin{(4x)}ln{(2)}}$

Explanation

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[a^u]=a^uduln(a)\&\text{Solve step-by-step:}\&f(x,y,z)=\frac{{(2^{(4y)}\cdot2^{(3z)}cos{(4x)})}}{2}\&f_{x}=-2\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}\&f_{xz}=-6\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}ln{(2)}\end{align*}$

Find the derivative of the function with respect to by taking the natural logarithm of both sides and then differentiating both sides with respect to .

$z=\frac{xy+2xy}{4y\sqrt{y^2+2x}}$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left(\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right )$

CORRECT

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x+2x}{xy+2xy} +\frac{4}{x}+\frac{4x+1}{2x^2+x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{3y}{xy+2xy} +\frac{4y+1}{2y^2+y} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x^2}{x^3y+xy} +\frac{5}{5y+x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x}{x^3y+x} +\frac{5y}{5y^2+x} \right )$

Explanation

$z=f(x,y)=\frac{xy+2xy}{4y\sqrt{y^2+2x}}$

Take the natural logarithm of both sides and expand the right-side using the properties of logarithms.

$\ln(z)=\ln\left(\frac{xy+2xy}{4y\sqrt{y^2+2x}}\right)$

Apply the rule for the logarithm of a quotient:

$\ln(z)=\ln\left(xy+2xy\right)-\ln\left(4y\sqrt{y^2+2x} \right)$

Apply the rule for the logarithm of a product:

$\ln(z)=\ln\left(xy+2xy\right)-\left[\ln(4y)+\ln\left(\sqrt{y^2+2x}\right) \right]$

Apply the rule for the logarithm of a quantity raised to a power:

$\ln(z)=\ln\left(xy+2xy\right)-\ln(4y)-\frac{1}{2}\ln\left(y^2+2x\right)$

Now differentiate both sides implicitly, remembering that is a multi-variable function of and .

$\frac{1}{z}\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}\left[\ln\left(xy+2xy\right)-\ln(4y)-\frac{1}{2}\ln\left(y^2+2x\right)\right]$

Proceeding with the differentiation on the right-side with respect to .

$\frac{1}{z}\frac{\partial z}{\partial y}=\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x}$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left[\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right ]$

Evaluate the limit:

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}$

$\frac{1}{6}$

CORRECT

Does not exist

$-\infty$

$\infty$

Explanation

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=0; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}=\frac{0^2+2*0+1}{0+6}=\frac{1}{6}$

Evaluate the limit:

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

$\frac{1}{4}$

CORRECT

$\infty$

$-\infty$

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):

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

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

$\begin{align*}&\text{y-axis: }0\&y=x:-20\end{align*}$

CORRECT

$\begin{align*}&\text{y-axis: }-380\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }220\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\&y=x:-\frac{5}{4}\end{align*}$

Explanation

$\begin{align*}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&\frac{(3(0)y^{2} + 17(0)^{2}y)}{(3(0)^{3} - 4y^{3})}=\frac{0}{-4y^{3}}\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{-4y^{3}}=0\&\text{We can make a similar substitution for }y=x:\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(3x(x)^{2} + 17x^{2}(x))}{(3x^{3} - 4(x)^{3})}=-20\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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

${45x^{2}y^{2}zcos{(y^{3})} - 8xcos{(x^{2})}cos{(4y)}sin{(4z)}}$

CORRECT

${-{(15x^{2}zsin{(y^{3})} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)})}\cdot{(8xcos{(x^{2})}cos{(4y)}sin{(4z)} - 45x^{2}y^{2}zcos{(y^{3})})}}$

${15x^{2}zsin{(y^{3})} - 4cos{(4y)}sin{(x^{2})}sin{(4z)} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)} + 15x^{3}y^{2}zcos{(y^{3})}}$

${-{(15x^{2}zsin{(y^{3})} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)})}\cdot{(4cos{(4y)}sin{(x^{2})}sin{(4z)} - 15x^{3}y^{2}zcos{(y^{3})})}}$

Explanation

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\&\text{simply treat any variable that you're not deriving the equation for as constant.}\&\text{Utilizing derivative rules:}\&d[uv]=udv+vdu\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\&\text{Solve step-by-step:}\&f(x,y,z)=5x^{3}zsin{(y^{3})} - sin{(x^{2})}sin{(4y)}sin{(4z)}\&f_{x}=15x^{2}zsin{(y^{3})} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)}\&f_{xy}=45x^{2}y^{2}zcos{(y^{3})} - 8xcos{(x^{2})}cos{(4y)}sin{(4z)}\end{align*}$

Return to subject

AI Study CoachYour personal study buddy

Instant TutoringConnect with a tutor now

Request a TutorGet matched with an expert

Live ClassesJoin interactive classes

Question of the DayDaily practice to build your skills

FlashcardsStudy and memorize key concepts

WorksheetsBuild custom practice quizzes

AI SolverStep-by-step problem solutions

Learn by ConceptMaster concepts step by step

Diagnostic TestsAssess your knowledge level

Practice TestsTest your skills with exam questions

GamesLearn through free games