# 8.2: This is why partial derivatives are so easy...

- Page ID
- 3397

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In your first calculus class you learned about derivatives. Suppose we have a function * f* that is a function of

*, which we can write as*

*x**(*

*f**). What is the derivative of*

*x**with respect to*

*f**?*

*x*\[\dfrac{d f(x)}{d x}\]

What about a new function that depends on two variables, * h*(

*)? This function could, for example, give the height*

*x,y**of mountainous terrain for each horizontal point (*

*h**). So what is the derivative of*

*x,y**with respect to*

*h**? One way we determine this derivative is to fix the value of*

*x**=*

*y**, which is the same as assuming that*

*y*_{1}*is a constant, and then take the ordinary derivative of*

*y**with respect to*

*h**. In a sense, we are taking a slice through the mountain in the*

*x**-direction at a fixed value of*

*x**=*

*y**. Thus,*

*y*_{1}\[\left(\frac{d h}{d x}\right)_{y=\text { constant }} \equiv \frac{\partial h}{\partial x}\]

This is called the **partial derivative** of * h* with respect to

*. It’s pretty easy to determine because we do not need to worry about how*

*x**might depend on*

*y**.*

*x*Exercise

Let \(h=(x-3)^{2} \cos (y)\) . What is the partial derivative of * h* with respect to

*?*

*x***Click for answer**-
\[\frac{\partial h}{\partial x}=\frac{\partial\left((x-3)^{2} \cos (y)\right)}{\partial x}=2(x-3) \cos (y) \nonumber\]

We can also find the partial derivative of * h* with respect to

*. Can you do this?*

*y*-
**Click for answer** -
\[\frac{\partial h}{\partial y}=\frac{\partial\left((x-3)^{2} \cos (y)\right)}{\partial y}=-(x-3)^{2} \sin (y) \nonumber\]

So you can see that the ∂h/∂x∂h/∂x may be different for each value of

and ∂h/∂y∂h/∂y may be different for each value of*y*. Thus, even if you are not entirely familiar with partial derivatives and their notation, you can see that they are no different from ordinary derivatives but you take the derivative for just of one variable at a time.*x*

Need more practice?

**Partial Derivatives**

**Click here for transcript of the Partial Derivatives Video**-
Hey everyone! Welcome back! Today we're going to be doing some Partial Derivative problems. The first one that we're going to do is f of (x, y, z) equals x squared times y cubed times z to the four, and all that partial derivatives means is that we're going to be taking the derivative for every single variable in the problem. So, since there are three variables, we're actually going to have to take the derivative three times, once for each variable, and its' going to be a separate equation four each one. So we'll go in order. The first one we'll look at is x and the notation for partial derivative looks like this. This weird squiggly thing f, the weird squiggly thing x, so f, you get from here, so this is, you know, h of (x, y, z) then this becomes an h and then x is the first variable we're going to... we're going to do. So, we'll go ahead and say... now, when we're taking and... and when... when we take the partial derivative here with x, we say that we're taking the partial derivative with respect to x, that's how people call it, then we'll take the partial derivative with respect to y then with respect to z. So, looking at x, the way that we take the partial derivative with respect to x while still having these other variables in the equation is we treat the other variables like they're constants. And, what I like to do, and you get... you'll get faster and faster at it in your head, but the way that I like to do that just to make it really obvious because sometimes it's hard to understand how to hold those constant especially when... when you're first starting out, I like to actually put a constant in there for those numbers and then simplify the equation and then take the partial derivative so I can see it. So... so, what I would do here, for example, we're talking about holding y and z constant, as if they we're a constant number like two or three, so let's go ahead and put two in for... for... for y and z here. If we did, we would have... we would have x squared times two cubed times two to the fourth, right, because we... we plugged in two for y and for z. Okay? So, since we're keeping this constant, this is how the equation would simplify. So if... if you... if you multiply this out, let's see, this would actually be x squared times eight and this would be times sixteen, so this would be eighty and forty eight, so this would be x squared times a hundred and twenty eight, right, if you simplify that. So, what would be if we... if we were taking the derivative of this normally, we would be looking at a hundred and twenty eight x squared. We would take the derivative of this and if we would multiply two times the coefficient which is a hundred and twenty eight so that would be, what? Two hundred and fifty six, so the derivative of this would be two hundred and fifty six x, right? So, what I'm hoping that you can see from this is that it's... its... it's exactly the same thing. We're going to hold these two things constant and they are going to be like a coefficient and... and this two hundred and fifty six stays. So, this is actually going to be an... Let's go ahead and write out the answer and then we'll compare them. You're going to multiply these two out in front here so it's going to be two x and then y cubed z to the four. That's going to be the answer for the partial derivative and I that you can see the relationship here. We multiplied the two on the x squared out in front just like we did here, we... we brought this two out in front, and we ended up with a single x just like we ended up with a single x here and we left y cubed and z to the fourth because they were absorbed into the coefficient here. They are like... because they're constants and they're multiplied together, they are part of the coefficient, they're like part of this two which is why... which is why they get left in... in this equation. Let's go ahead and do y so that we see in another example and hopefully we'll start to understand. So, when we... when we take the partial derivative with respect to y as you might expect, it's going to be a partial derivative of f with respect to y, just like we did for x here. So now, with y, we're going to actually be holding... Oh, I hope you guys can't hear that fire truck. So with... with y, we're going to be keeping x and z constant so they're going to be like the coefficient as well, you could plug in numbers for them and... and go through the same exercise. But, they're like the coefficient so they're going to stay exactly the same because they're... they're multiplied here with the y so we're not even going to touch them. Remember, we didn't touched y cubed, we didn't touch z to the fourth so x and z, this time, are going to stay as well. So, all we're really looking at is... is the y and we're going to... we're going to do the same thing we did with x, take the derivative of y. So we're going to get that three out in front and then y squared, right, three y squared is the derivative of y cubed. So we took the derivative and then the x squared and the z to the fourth are just going to stay. So, that's the derivative with respect to y. I left the space because when you take the... the partial derivative, you always like to keep the variables in alphabetical order. So, I could have written three y squared x squared z to the fourth but we like to always keep them x, y, z in order. So, we'll go ahead and... and do the same thing here for z. So it's going to be the partial derivative of f with respect to z and we will go ahead and leave x squared and y cubed. We're not touching them because they're like part of the coefficient, they stay. So we'll go ahead and say x squared y cubed and then we... we take the derivative of z here. So the derivative... Of course, we subtract one from the exponent, so four minus one is three and the four gets multiply out in front so it comes out here. So our answer with respect to z is actually four x squared y cubed z cubed. And, your final answer is... is a three part answer if you're asked to take the... the... the derivative of this function or the partial derivative. Because there are three variables, you need each one of these equations and you would want to write all three of these down on the homework or on your test because... because your answer is actually all three of these. So, there you have it.