This section covers:
Differentials, Linear Approximation and Error Propagation are more applications of Differential Calculus.
Differentials
Think of differentials of picking apart the “fraction” \(\displaystyle \frac{{dy}}{{dx}}\) we learned to use when differentiating a function.
We learned that the derivative or rate of change of a function can be written as \(\displaystyle \frac{{dy}}{{dx}}={f}’\left( x \right)\), where \(dy\) is an infinitely small change in \(y\), and \(dx\) (or \(\Delta x\)) is an infinitely small change in \(x\). It turns out that if \(f\left( x \right)\) is a function that is differentiable on an open interval containing \(x\), and the differential of \(x\) (\(dx\)) is a nonzero real number, then \(dy={f}’\left( x \right)dx\) (see how we just multiplied both sides by \(dx\))? And I won’t get into this at this point, but the differential of \(y\) can be used to approximate the change in \(y\), so \(\Delta y\approx dy\). (I always forget this, but remember that \(\Delta y=f\left( {x+\Delta x} \right)f\left( x \right)\).)
Calculating Differentials
We learned differentiation rules earlier, and these apply to differentials too. These look familiar, right? We’ll see that we’ll need to use the differential produce rule in the problem here.
Here are the differential formulas:
If \(u\) and \(v\) are differentiable functions of \(x\):
Constant Multiple: \(d\left[ {cu} \right]=c\,du\)
Sum/Difference: \(d\left[ {u\pm v} \right]=du\pm dv\)
Product: \(d\left[ {uv} \right]=u\,dv+v\,du\)
Quotient: \(\displaystyle d\left[ {\frac{u}{v}} \right]=\frac{{v\,duudv}}{{{{v}^{2}}}}\)
Linear Approximation
We can use differentials to perform linear approximations of functions (we did this here with tangent line approximation) with this formula that looks similar to a pointslope formula (remember that the derivative is a slope): \(y{{y}_{0}}={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\), or \(f\left( x \right)f\left( {{{x}_{0}}} \right)={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\), which means \(f\left( x \right)=f\left( {{{x}_{0}}} \right)+{f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\). And remember that the variables with subscript “0” are the “old” values. Think of the equation as the “new \(y\)” equals the “old \(y\)” plus the derivative at the “old \(x\)” times the difference between the “new \(x\)” and the “old \(x\)”.
(And remember that we do these types of problems so we can “appreciate math” the way those used calculus before calculators and computers.)
Note: Another way we can think of differentials is using this formula; some teachers prefer this way: \(\displaystyle \frac{{dy}}{{dx}}={f}’\left( x \right);\,\,dy={f}’\left( x \right)dx\) (this makes sense, right? A “slope” is a “slope”). Then once we get \(dy\), we just add it back to the original \(y\) to get the approximation. This is also shown in the fourth problem below.
Here are some examples in both finding differentials and finding approximations of functions:
Problem  Solution  
Find the value of \(\boldsymbol {dy}\) and \(\boldsymbol {\Delta y}\) for \(x=4\) and \(\Delta x=.1\).
(Remember that \(\Delta y=f\left( {x+\Delta x} \right)f\left( x \right)\))
(The answers are close since \(\Delta x\) is small) 
First, find \(dy\) by differentiating:
\(\displaystyle y={{x}^{2}}1;\,\,\,\,\frac{{dy}}{{dx}}=2x;\,\,\,dy=\,2x\cdot dx\) When \(x=4\) and \(\Delta x=.1\), \(\displaystyle dy=2\left( 4 \right)\,\cdot .1=.8\).
\(\displaystyle \begin{align}\Delta y&=f\left( {x+\Delta x} \right)f\left( x \right)\\&=f\left( {4+.1} \right)f\left( 4 \right)\\&=\left( {{{{4.1}}^{2}}1} \right)\left( {{{4}^{2}}1} \right)=.81\end{align}\) 

Find the differential \(dy\) for: \(y=4\cos \left( {2x} \right)8{{x}^{3}}\) 
\(\displaystyle \begin{align}y&=4\cos \left( {2x} \right)8{{x}^{3}}\\\frac{{dy}}{{dx}}&=4\cdot \sin \left( {2x} \right)\cdot 224{{x}^{2}}\\dy&=\left( {8\sin \left( {2x} \right)24{{x}^{2}}} \right)dx\end{align}\)  
Use differentials and the graph of \({f}’\left( x \right)\) (derivative) to approximate \(f\left( {3.2} \right)\), given that \(f\left( 3 \right)=5\).

Use this formula: \(y{{y}_{0}}={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\) (I like to use this formula since it looks like pointslope. Remember that the variables with subscript 0 are the “original” or “old” values). Note that this is a variation of the formula \(f\left( x \right)=f\left( {{{x}_{0}}} \right)+{f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\). We see that:
Then we have \(y{{y}_{0}}={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\), or \(y5=2.25\left( {3.23} \right)\). So \(y=2.25\left( {3.23} \right)+5=5.45\). 

Use differentials to approximate: \(\sqrt{{15}}\)
Alternative way to solve without pointslope formula. Use 16 for \(x\), 4 for \({{y}_{0}}\), \(15–16=– 1\) for \(dx\): \(\displaystyle \begin{align}y&=\sqrt{x}={{x}^{{\frac{1}{2}}}}\\\frac{{dy}}{{dx}}&=\frac{1}{2}{{x}^{{\frac{1}{2}}}}\\dy&=\frac{1}{2}{{x}^{{\frac{1}{2}}}}dx\\dy&=\frac{1}{2}{{\left( {16} \right)}^{{\frac{1}{2}}}}\left( {1} \right)=\frac{1}{8}\end{align}\)
\(\displaystyle y={{y}_{0}}+dy=4+\frac{1}{8}=3.875\) 
Use this formula: \(y{{y}_{0}}={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\) The function is \(y=\sqrt{x}={{x}^{{\frac{1}{2}}}}\), so we have \(\displaystyle {f}’\left( x \right)=\frac{1}{2}{{x}^{{\frac{1}{2}}}}\). Now the trick here is to find an easier value in the function so we can solve it without a calculator; let’s use \(\sqrt{{16}}=4\). Now we have:
Then we have \(y{{y}_{0}}={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\), or \(\displaystyle y4=.125\left( {1516} \right)\). Then \(\displaystyle y=.125\left( {1516} \right)+4=3.875\).
Compare this to what you get on your calculator. Pretty cool! 

Use differentials to approximate:
\(\displaystyle \sin \left( 3 \right)\)

Use this formula: \(y{{y}_{0}}={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\) The function is \(y=\sin \left( x \right)\), so we have \({f}’\left( x \right)=\cos \left( x \right)\). Now the trick here is to find an easier value in the function so we can solve it without a calculator; let’s use \(\sin \left( \pi \right)\,\,(\pi \approx 3.14)\). Now we have:
Then we have \(y{{y}_{0}}={f}’\left( {{{x}_{0}}} \right)\left( {x{{x}_{0}}} \right)\), or \(\displaystyle y0=1\left( {3\pi } \right)\). Then \(\displaystyle y=\pi 3=.14112\). 
Error Propagation
We can also use differentials in Physics to estimate errors, say in physical measuring devices. In these problems, we’ll typically take a derivative, and use the “\(dx\)” or “\(dy\)” part of the derivative as the error. Then, to get percent error, we’ll divide the error by the total amount and multiply by 100.
The other thing to remember is that when we are solving for an error, it can go either way, so we typically express our answers with a “\(\pm \)”.
We’ll attack these problems the same way we did with related rates problems: write down what we know, what we need, and how we relate the variables.
Here are some problems:
Error Propagation Problem  Solution 
The volume of a cube is 125 in^{3}.
If the volume measurement is known to be correct to within 2.5 in^{3}, estimate the error in the measurement of a side of the cube. 
We can attack this like a related rates problem: write down what we know, what we need, and how we relate the variables.
We have \(V=125\), and \(dV=2.5\) (remember that the error is the “\(dx\)” part of the equation). We want the error in the side of the cube, so we want \(ds\). Now let’s relate the variables and differentiate with respect to \(s\): \(\displaystyle V={{s}^{3}};\,\,\,\,\,\frac{{dV}}{{ds}}=3{{s}^{2}};\,\,\,\,\,dV=3{{s}^{2}}ds\) Substitute and solve for \(ds\). Note that since we know the volume (\(V\)) is 125, we know a side (\(s\)) is \(\sqrt[3]{{125}}=5\): \(\displaystyle dV=3{{s}^{2}}ds;\,\,\,\,\,2.5=3{{\left( 5 \right)}^{2}}ds;\,\,\,\,ds=\frac{{2.5}}{{3{{{\left( 5 \right)}}^{2}}}}=\pm \,\,.0333\) The error in the measurement of a side of the cube is ±.0333 in. 
The radius of a sphere is measured to be 5 mm. If this measurement is correct to within .05 mm,
a) Estimate the propagated error in the surface area of the sphere.
b) Estimate the propagated error in the volume of the sphere.
c) Estimate the percent error of the volume of the sphere. 
We have \(r=5\), and \(dr=.05\) (remember that the error is the “\(dx\)” part of the equation). We want the both the error in the surface area (\(dA\)) and the error in the volume (\(dV\)). We have to remember the equations from Geometry.
a) Let’s relate surface area and radius, and differentiate with respect to \(r\): \(\displaystyle A=4\pi {{r}^{2}};\,\,\,\,\,\frac{{dA}}{{dr}}=8\pi r;\,\,\,\,\,dA=\left( {8\pi r} \right)dr\) We want \(dA\) (error in the surface area): \(\displaystyle dA=\left( {8\pi r} \right)dr;\,\,\,\,\,dA=\left( {8\pi \cdot 5} \right).05\approx \pm \,\,6.283\) The error in the measurement of the surface area of the sphere is ±6.283 mm^{2}.
b) Let’s relate volume and radius, and differentiate with respect to \(r\): \(\displaystyle V=\frac{4}{3}\pi {{r}^{3}};\,\,\,\,\,\frac{{dV}}{{dr}}=4\pi {{r}^{2}};\,\,\,\,\,dV=\left( {4\pi {{r}^{2}}} \right)dr\) We want \(dV\) (error in the volume): \(\displaystyle dV=\left( {4\pi {{r}^{2}}} \right)dr;\,\,\,\,\,dV=\left( {4\pi \cdot {{5}^{2}}} \right).05\approx \pm \,\,15.708\) The error in the measurement of the volume of the sphere is ±15.708 mm^{3}. c) To get percent error: \(\displaystyle \begin{align}\text{Percent Error}&=\frac{{\text{ Error}}}{{\text{Volume}}}\cdot 100=\frac{{dV}}{V}\cdot 100\\&=\frac{{15.708}}{{\frac{4}{3}\pi {{r}^{3}}}}=\frac{{15.708}}{{\frac{4}{3}\pi {{{\left( 5 \right)}}^{3}}}}\approx .03\cdot 100=3\end{align}\) The percent error in the measurement of the volume of the sphere is 3%.

Here are a few more that are a bit more difficult; for the first below, we need to use the Differentials Product Rule:
Error Propagation Problem  Solution 
The measurement of the base and altitude of a triangle are 20 cm and 30 cm, respectively. The possible error in each measurement is .3 cm.
a) Estimate the possible propagated error in computing the area of the triangle. b) Approximate the percent error in computing this area. 
This one’s a little trickier since we have two variables with error. We have \(b=20,\,\,h=30,\,\,db=.3,\,\,\,dh=.3\) (remember that the error is the “\(dx\)” part of the equation).
We want the error in the area of the triangle (\(dA\)), so we have to relate the two perpendicular sides of the triangle to the area: \(\displaystyle A=\frac{1}{2}bh\). We’ll be using the differential product rule, \(d\left[ {uv} \right]=u\,dv+v\,du\). We can also notice that \(\require {cancel} \displaystyle A=\frac{1}{2}bh;\,\,\,\frac{{dA}}{{dx}}=\frac{1}{2}\left( {b\cdot \frac{{dh}}{{dx}}+h\cdot \frac{{db}}{{dx}}} \right);\,\,dA=\frac{1}{2}\left( {b\cdot \frac{{dh}}{{\cancel{{dx}}}}+h\cdot \frac{{db}}{{\cancel{{dx}}}}} \right)\cancel{{dx}}\), so we have \(\displaystyle dA=\frac{1}{2}\left( {b\cdot dh+h\cdot db} \right)\). Now substitute and solve for \(dA\): \(\displaystyle dA=\frac{1}{2}\left( {b\cdot dh+h\cdot db} \right)=\frac{1}{2}\left( {20\times .3+30\times .3} \right)=7.5\)
a) The error in the measurement of the area of the triangle is ±7.5 cm^{2}. b) The percent error in the area is \(\displaystyle \frac{{\text{error in area}}}{{\text{total area}}}\cdot 100=\frac{{dA}}{A}\cdot 100=\frac{{7.5}}{{\frac{1}{2}\cdot 20\cdot 30}}\times 100=2.5\%\). 
How much variation \(dr\) in the radius of a coin can be tolerated if the volume of the coin is to be within \(\displaystyle \frac{1}{{1000}}\) of their ideal volume?  It’s difficult to know how to start this problem, but let’s start out by relating radius with volume by using the volume of a cylinder equation.
We want \(dr\), and have \(dV\) (\(\displaystyle \frac{1}{{1000}}\)), so this will fit into using this equation and differentiating: \(\displaystyle V=\pi {{r}^{2}}h;\,\,\,\,\,\,\frac{{dV}}{{dr}}=2\pi rh;\,\,\,\,\,\,dV=2\pi rh\,dr\) Since we know the volume of the coin is to be within \(\displaystyle \frac{1}{{1000}}\) of volume of its total volume, we can interpret \(dV\) to be this error, so \(\displaystyle \left {dV} \right\le .001V\). Now let’s solve for \(dr\): \(\displaystyle \left {dV} \right=\left {2\pi rh\,dr} \right\le .001V,\,\,\,\,\text{or}\,\,\,\,\left {2\pi rh\,dr} \right\le .001\left( {\pi {{r}^{2}}h} \right)\). Therefore, \(\displaystyle \,\left {dr} \right\le \frac{{.001\left( {\cancel{\pi }{{r}^{{\cancel{2}}}}\cancel{h}} \right)}}{{2\cancel{\pi }\cancel{r}h}}\), or \(\displaystyle \,\left {dr} \right\le .0005r\). The variation of the radius should not exceed .05% of the ideal radius. Tricky!

Learn these rules and practice, practice, practice!
On to Exponential and Logarithmic Differentiation — you are ready!