## MATH1111 Quizzes

Computing Partial Derivatives Algebraically Quiz
Web resources available Questions

This quiz tests the work covered in the lecture on Computing Partial Derivatives and corresponds to Section 14.2 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 2. This quiz was the same as the Section 1 quiz at 14/12/05.

There are some good examples of partial derivatives and some easy exercises at http://www.analyzemath.com/calculus/multivariable/partial˙derivatives.html and an explanation at http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd3.html and more exercises at http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd4.html.

Suppose $f\left(x,y\right)=3{x}^{2}+2x{y}^{3}+4{y}^{2}\phantom{\rule{0.3em}{0ex}}.$ Which one of the following statements is correct? Exactly one option must be correct)
 a) ${f}_{x}\left(x,y\right)=6x+6{y}^{2}+8y$ and ${f}_{y}\left(x,y\right)=3{x}^{2}+6xy+8y$ b) ${f}_{x}\left(x,y\right)=6x+2{y}^{3}+4{y}^{2}$ and ${f}_{y}\left(x,y\right)=6x+6x{y}^{2}+8y$ c) ${f}_{x}\left(x,y\right)=6x+2{y}^{3}$ and ${f}_{y}\left(x,y\right)=6x{y}^{2}+8y$ d) ${f}_{x}\left(x,y\right)=6x+6x{y}^{2}$ and ${f}_{y}\left(x,y\right)=2{y}^{3}+8y$

Choice (a) is incorrect
Try again, remember to treat the $y$’s as a constant when differentiating with respect to $x$ and to treat the $x$’s as a constant when differentiating with respect to $y\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, remember to treat the $y$’s as a constant when differentiating with respect to $x$ and to treat the $x$’s as a constant when differentiating with respect to $y\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is correct!
Choice (d) is incorrect
Try again, remember to treat the $y$’s as a constant when differentiating with respect to $x$ and to treat the $x$’s as a constant when differentiating with respect to $y\phantom{\rule{0.3em}{0ex}}.$
Suppose $f\left(x,y\right)={x}^{3}{e}^{xy}\phantom{\rule{0.3em}{0ex}}.$ Which one of the statements is correct? Exactly one option must be correct)
 a) $\frac{\partial f}{\partial x}=3{x}^{2}{e}^{xy}+{x}^{3}y{e}^{xy}$ and $\frac{\partial f}{\partial y}={x}^{4}{e}^{xy}$ b) $\frac{\partial f}{\partial x}=3{x}^{2}y{e}^{xy}$ and $\frac{\partial f}{\partial y}=3{x}^{3}{e}^{xy}$ c) $\frac{\partial f}{\partial x}=3{x}^{2}{e}^{xy}+{x}^{4}{e}^{xy}$ and $\frac{\partial f}{\partial y}={x}^{3}y{e}^{xy}$ d) $\frac{\partial f}{\partial x}=3{x}^{2}{e}^{xy}$ and $\frac{\partial f}{\partial y}={x}^{3}{e}^{xy}$

Choice (a) is correct!
You have correctly used the product rule when differentiating with respect to $x$ and $\frac{\partial f}{\partial y}={x}^{3}x{e}^{xy}={x}^{4}{e}^{xy}$
Choice (b) is incorrect
Try again, you must use the product rule to differentiate with respect to $x$ and regard ${x}^{3}$ as a constant when you differentiate with respect to $y\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, you have not differentiated ${e}^{xy}$ correctly with respect to either variable.
Choice (d) is incorrect
Try again, you must use the product rule to differential with respect to $x$ and regard ${x}^{3}$ as a constant when you differentiate with respect to $y\phantom{\rule{0.3em}{0ex}}.$
Consider $z={\left({x}^{2}y+3x{y}^{3}\right)}^{4}\phantom{\rule{0.3em}{0ex}}.$ Which one of the following statements is correct? Exactly one option must be correct)
 a) $\frac{\partial z}{\partial x}=4{\left({x}^{2}y-3x{y}^{3}\right)}^{3}\left(2x+9{y}^{2}\right)$ and $\frac{\partial z}{\partial y}=4{\left({x}^{2}y-3x{y}^{3}\right)}^{3}\left(2x+9{y}^{2}\right)$ b) $\frac{\partial z}{\partial x}=4{\left({x}^{2}y-3x{y}^{3}\right)}^{3}\left(2xy+3{y}^{3}\right)$ and $\frac{\partial z}{\partial y}=4{\left({x}^{2}y-3x{y}^{3}\right)}^{3}\left({x}^{2}+9x{y}^{2}\right)$ c) $\frac{\partial z}{\partial x}=4{\left(2xy+3{y}^{3}\right)}^{3}$ and $\frac{\partial z}{\partial y}=4{\left({x}^{2}+9x{y}^{2}\right)}^{3}$ d) $\frac{\partial z}{\partial x}=4\left({x}^{2}y-3x{y}^{3}\right){\left(2xy+3{y}^{3}\right)}^{3}$ and $\frac{\partial z}{\partial y}=4\left({x}^{2}y-3x{y}^{3}\right){\left({x}^{2}+9x{y}^{2}\right)}^{3}$

Choice (a) is incorrect
Try again, you are not differentiating the internal function correctly.
Choice (b) is correct!
You have used the chain rule correctly.
Choice (c) is incorrect
Try again, you have not used the chain rule correctly.
Choice (d) is incorrect
Try again, you are not using the chain rule correctly.
Suppose $f\left(x,y\right)=\frac{sinxy}{{x}^{2}{y}^{3}}\phantom{\rule{0.3em}{0ex}}.$ Which of one the following statements is correct? Reduce your answer to a fraction in lowest terms. Exactly one option must be correct)
 a) ${f}_{x}\left(x,y\right)=\frac{ycosxy}{2x{y}^{3}}$ and ${f}_{y}\left(x,y\right)=\frac{xcosxy}{3{x}^{2}{y}^{3}}$ b) ${f}_{x}\left(x,y\right)=-\frac{xycosxy+2sinxy}{{\left(xy\right)}^{3}}$ and ${f}_{y}\left(x,y\right)=-\frac{xycosxy+3sinxy}{{x}^{2}{y}^{4}}$ c) ${f}_{x}\left(x,y\right)=\frac{{x}^{2}cosxy-2sinxy}{{\left(xy\right)}^{3}}$ and ${f}_{y}\left(x,y\right)=\frac{{y}^{2}cosxy-3sinxy}{{x}^{2}{y}^{4}}$ d) ${f}_{x}\left(x,y\right)=\frac{xycosxy-2sinxy}{{\left(xy\right)}^{3}}$ and ${f}_{y}\left(x,y\right)=\frac{xycosxy-3sinxy}{{x}^{2}{y}^{4}}$

Choice (a) is incorrect
Try again, using the quotient rule.
Choice (b) is incorrect
Try again, recall $\frac{d\left(sinx\right)}{dx}=cosx\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, you have not differentiated $sinxy$ correctly.
Choice (d) is correct!
 ${f}_{x}\left(x,y\right)$ = $\frac{\left(ycosxy\right)\left({x}^{2}{y}^{3}\right)-sinxy\left(2x{y}^{3}\right)}{{\left({x}^{2}{y}^{3}\right)}^{2}}$ = $\frac{{x}^{2}{y}^{4}cosxy-2x{y}^{3}sinxy}{{x}^{4}{y}^{6}}$ = $\frac{x{y}^{3}\left(xycosxy-2sinxy\right)}{{x}^{4}{y}^{6}}$ = $\frac{xycosxy-2sinxy}{{\left(xy\right)}^{3}}$
and
 ${f}_{y}\left(x,y\right)$ = $\frac{\left(xcosxy\right)\left({x}^{2}{y}^{3}\right)-sinxy\left({x}^{2}3{y}^{2}\right)}{{\left({x}^{2}{y}^{3}\right)}^{2}}$ = $\frac{{x}^{3}{y}^{3}cosxy-3{x}^{2}{y}^{2}sinxy}{{x}^{4}{y}^{6}}$ = $\frac{{x}^{2}{y}^{2}\left(xycosxy-3sinxy\right)}{{x}^{4}{y}^{6}}$ = $\frac{xycosxy-3sinxy}{{x}^{2}{y}^{4}}$