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. So 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 \(\Delta y\approx dy\).
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:
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. So 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.)
Here are some examples in both finding differentials and finding approximations of functions:
Problem  Solution  
Find the differential \(dy\) for: \(y=4\cos \left( {2x} \right)8{{x}^{3}}\) 
\(\displaystyle \begin{array}{l}y=4\cos \left( {2x} \right)8{{x}^{3}}\\\frac{{dy}}{{dx}}=4\cdot \sin \left( {2x} \right)\cdot 224{{x}^{2}}\\dy=\left( {8\sin \left( {2x} \right)24{{x}^{2}}} \right)dx\end{array}\)  
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 “±”.
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:
Here are a few more that are a bit more difficult; for the first below, we need to use the Differentials Product Rule:
Learn these rules and practice, practice, practice!
On to Exponential and Logarithmic Differentiation — you are ready!