Differential of a Function
Solved Problems
Example 9.
Find the differential of the function \[y = x{e^{{x^2}}}.\]
Solution.
We calculate the derivative using the product and chain rule:
\[y^\prime = \left( {x{e^{{x^2}}}} \right)^\prime = x^\prime{e^{{x^2}}} + x\left( {{e^{{x^2}}}} \right)^\prime = 1 \cdot {e^{{x^2}}} + x \cdot {e^{{x^2}}} \cdot \left( {{x^2}} \right)^\prime = {e^{{x^2}}} + x \cdot {e^{{x^2}}} \cdot 2x = {e^{{x^2}}} + 2{x^2}{e^{{x^2}}} = {e^{{x^2}}}\left( {1 + 2{x^2}} \right).\]
The differential of the function is written as
\[dy = y^\prime dx = {e^{{x^2}}}\left( {1 + 2{x^2}} \right)dx.\]
Example 10.
Find the differential of the function \[y = {\sin ^2}3x.\]
Solution.
First we determine the derivative using the chain rule:
\[y^\prime = \left( {{{\sin }^2}3x} \right)^\prime = 2\sin 3x \cdot \left( {\sin 3x} \right)^\prime = 3 \cdot 2\sin 3x\cos 3x.\]
Apply the double angle identity
\[\sin 2\alpha = 2\sin \alpha \cos \alpha .\]
Hence
\[y^\prime = 3 \cdot 2\sin 3x\cos 3x = 3\sin 6x.\]
The differential of a function is written in the form
\[dy = y^\prime dx.\]
Then
\[dy = 3\sin 6xdx.\]
Example 11.
Find the differential of \[y = x\sin {\frac{{\pi x}}{2}}\] at the point \(x = {\frac{1}{2}}\) when \(dx = 0,01.\)
Solution.
\[dy = f'\left( y \right)dx = {\left( {x\sin \frac{{\pi x}}{2}} \right)^\prime }dx = \left( {1 \cdot \sin \frac{{\pi x}}{2} + x \cdot \cos \frac{{\pi x}}{2} \cdot \frac{\pi }{2}} \right)dx = \left( {\sin \frac{{\pi x}}{2} + \frac{{\pi x}}{2}\cos \frac{{\pi x}}{2}} \right)dx.\]
Substitute the values of \(x\) and \(dx\) and calculate the differential \(dy:\)
\[dy = \left( {\sin \frac{{\pi \cdot \frac{1}{2}}}{2} + \frac{{\pi \cdot \frac{1}{2}}}{2}\cos \frac{{\pi \cdot \frac{1}{2}}}{2}} \right) \cdot 0,01 = \left( {\sin \frac{\pi }{4} + \frac{\pi }{4}\cos \frac{\pi }{4}} \right) \cdot 0,01 = \left( {\frac{{\sqrt 2 }}{2} + \frac{\pi }{4}\frac{{\sqrt 2 }}{2}} \right) \cdot 0,01 = \frac{{\sqrt 2 }}{{200}}\left( {1 + \frac{\pi }{4}} \right) \approx 0,0126.\]
Example 12.
Find the differential of the function \[y = \sqrt {{x^3} + 4x} \] at \(x = 2.\)
Solution.
Differentiate the given function:
\[y^\prime = \left( {\sqrt {{x^3} + 4x} } \right)^\prime = \frac{1}{{2\sqrt {{x^3} + 4x} }} \cdot \left( {{x^3} + 4x} \right)^\prime = \frac{{3{x^2} + 4}}{{2\sqrt {{x^3} + 4x} }}.\]
At the point \(x = 2,\) the derivative is equal to
\[y^\prime\left( 2 \right) = \frac{{3 \cdot {2^2} + 4}}{{2\sqrt {{2^3} + 4 \cdot 2} }} = \frac{{16}}{{2\sqrt {16} }} = 2.\]
Hence, the differential of the function at this point is
\[dy = y^\prime\left( 2 \right)dx = 2dx.\]
Example 13.
Use differential to approximate the change in \[y = \frac{1}{{\sin x}}\] as \(x\) changes from \(\frac{\pi }{4}\) to \(\frac{3\pi }{10}.\)
Solution.
We calculate the differential of the function by the formula
\[dy = y^\prime dx = y^\prime\left( {\frac{\pi }{4}} \right)dx.\]
The derivative is given by
\[y^\prime = \left( {\frac{1}{{\sin x}}} \right)^\prime = - \frac{1}{{{{\left( {\sin x} \right)}^2}}} \cdot \left( {\sin x} \right)^\prime = - \frac{{\cos x}}{{{{\sin }^2}x}}.\]
At the initial point \(x = \frac{\pi }{4},\) the derivative is equal to
\[y^\prime\left( {\frac{\pi }{4}} \right) = - \frac{{\cos \frac{\pi }{4}}}{{{{\sin }^2}\frac{\pi }{4}}} = - \frac{{\frac{{\sqrt 2 }}{2}}}{{{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}} = - \frac{2}{{\sqrt 2 }} = - \sqrt 2 .\]
Calculate the differential of the independent variable:
\[dx = \frac{{3\pi }}{{10}} - \frac{\pi }{4} = \frac{{6\pi - 5\pi }}{{20}} = \frac{\pi }{{20}}.\]
Then the differential \(dy\) is
\[dy = y^\prime\left( {\frac{\pi }{4}} \right)dx = - \sqrt 2 \cdot \frac{\pi }{{20}} = - \frac{{\sqrt 2 \pi }}{{20}}.\]
The approximate value of the function at \({x = \frac{{3\pi }}{{10}}}\) is equal to
\[y\left( {\frac{{3\pi }}{{10}}} \right) \approx y\left( {\frac{\pi }{4}} \right) + dy = \frac{1}{{\sin \frac{\pi }{4}}} - \frac{{\sqrt 2 \pi }}{{20}} = \frac{1}{{\frac{{\sqrt 2 }}{2}}} - \frac{{\sqrt 2 \pi }}{{20}} = \sqrt 2 - \frac{{\sqrt 2 \pi }}{{20}} = \frac{{\sqrt 2 }}{{20}}\left( {20 - \pi } \right).\]
Example 14.
Calculate the increment and differential of the function \[y = {\frac{{x + 2}}{{x + 1}}}\] at the point \(x = 0\) when \(\Delta x = 0,1.\)
Solution.
First we compute the increment of the function:
\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = \frac{{x + \Delta x + 2}}{{x + \Delta x + 1}} - \frac{{x + 2}}{{x + 1}} = \frac{{0,1 + 2}}{{0,1 + 1}} - \frac{2}{1} = 1,9091 - 2 \approx - 0,0909.\]
At the same values of \(x\) and \(\Delta x\), the differential of the function is equal to
\[\require{cancel} dy = f'\left( x \right)\Delta x = {\left( {\frac{{x + 2}}{{x + 1}}} \right)^\prime }\Delta x = \frac{{\cancel{x} + 1 - \cancel{x} - 2}}{{{{\left( {x + 1} \right)}^2}}}\Delta x = - \frac{1}{{{{\left( {x + 1} \right)}^2}}}\Delta x = - \frac{1}{{{1^2}}} \cdot 0,1 = - 0,1.\]
Example 15.
Find the differential of the function \[y = \ln \tan 2x\] at \(x = \frac{\pi }{8}.\)
Solution.
Take the derivative of the function:
\[y^\prime = \left( {\ln \tan 2x} \right)^\prime = \frac{1}{{\tan 2x}} \cdot \left( {\tan 2x} \right)^\prime = \frac{1}{{\tan 2x}} \cdot \frac{1}{{{{\cos }^2}2x}} \cdot \left( {2x} \right)^\prime = \frac{2}{{\tan 2x{{\cos }^2}2x}} = \frac{2}{{\frac{{\sin 2x}}{{\cos 2x}} \cdot {{\cos }^2}2x}} = \frac{2}{{\sin 2x\cos 2x}} = \frac{4}{{2\sin 2x\cos 2x}} = \frac{4}{{\sin 4x}}.\]
At the point \(x = \frac{\pi }{8},\) we have
\[y^\prime\left( {\frac{\pi }{8}} \right) = \frac{4}{{\sin \left( {4 \cdot \frac{\pi }{8}} \right)}} = \frac{4}{{\sin \frac{\pi }{2}}} = \frac{4}{1} = 4.\]
Then
\[dy = y^\prime\left( {\frac{\pi }{8}} \right)dx = 4dx.\]
Example 16.
Calculate the differential of the function \[y = \frac{{2x}}{{{x^2} - 3}}\] at the point \(x = 3\) when \(dx = - \frac{3}{{15}}.\)
Solution.
Take the derivative:
\[y^\prime = \left( {\frac{{2x}}{{{x^2} - 3}}} \right)^\prime = \frac{{\left( {2x} \right)^\prime\left( {{x^2} - 3} \right) - 2x\left( {{x^2} - 3} \right)^\prime}}{{{{\left( {{x^2} - 3} \right)}^2}}} = \frac{{2 \cdot \left( {{x^2} - 3} \right) - 2x \cdot 2x}}{{{{\left( {{x^2} - 3} \right)}^2}}} = \frac{{2{x^2} - 6 - 4{x^2}}}{{{{\left( {{x^2} - 3} \right)}^2}}} = - \frac{{2{x^2} + 6}}{{{{\left( {{x^2} - 3} \right)}^2}}}.\]
At \(x = 3,\) the derivative is equal to
\[y^\prime\left( 3 \right) = - \frac{{2 \cdot {3^2} + 6}}{{{{\left( {{3^2} - 3} \right)}^2}}} = - \frac{{24}}{{36}} = - \frac{2}{3}.\]
So the differential \(dy\) at this point is
\[dy = y^\prime\left( 3 \right)dx = - \frac{2}{3} \cdot \left( { - \frac{3}{{15}}} \right) = \frac{2}{{15}}.\]
Example 17.
Find the differential of the function \[y = {\frac{1}{{\sqrt {{u^2} + {v^2}} }}},\] where \(u\) and \(v\) are differentiable functions of the variable \(x.\)
Solution.
Using the differentiation rules, we obtain:
\[dy = d\left( {\frac{1}{{\sqrt {{u^2} + {v^2}} }}} \right) = d\left[ {{{\left( {{u^2} + {v^2}} \right)}^{ - \frac{1}{2}}}} \right] = - \frac{1}{2}{\left( {{u^2} + {v^2}} \right)^{ - \frac{3}{2}}} d\left( {{u^2} + {v^2}} \right) = - \frac{{2udu + 2vdv}}{{2\sqrt {{{\left( {{u^2} + {v^2}} \right)}^3}} }} = - \frac{{udu + vdv}}{{\sqrt {{{\left( {{u^2} + {v^2}} \right)}^3}} }}.\]
Example 18.
Find the differential of the function \[y = \arcsin {\frac{u}{v}},\] where \(u\) and \(v\) are differentiable functions of \(x.\)
Solution.
\[dy = d\left( {\arcsin \frac{u}{v}} \right) = \frac{1}{{\sqrt {1 - {{\left( {\frac{u}{v}} \right)}^2}} }}d\left( {\frac{u}{v}} \right) = \frac{1}{{\sqrt {1 - {{\left( {\frac{u}{v}} \right)}^2}} }} \cdot \frac{{vdu - udv}}{{{v^2}}} = \frac{{\left| v \right|}}{{\sqrt {{v^2} - {u^2}} }} \cdot \frac{{vdu - udv}}{{{{\left| v \right|}^2}}} = \frac{{vdu - udv}}{{\left| v \right|\sqrt {{v^2} - {u^2}} }},\]
where \({v^2} \gt {u^2},\) \(v \ne 0.\)
Example 19.
The function \(y\left( x \right)\) is given implicitly by the equation \[{y^3} - 3xy + {x^3} = 3.\] Find its differential at the point \(\left( {2,1} \right).\)
Determine the derivative of the implicit function. Differentiating both sides with respect to \(x,\) we obtain:
\[\left( {{y^3} - 3xy + {x^3}} \right)^\prime = \left( 3 \right)^\prime,\;\;\Rightarrow 3{y^2}y' - \left( {3y + 3xy'} \right) + 3{x^2} = 0,\;\; \Rightarrow 3{y^2}y' - 3y - 3xy' + 3{x^2} = 0,\;\; \Rightarrow \left( {{y^2} - x} \right)y' = y - {x^2},\;\; \Rightarrow y' = \frac{{y - {x^2}}}{{{y^2} - x}}.\]
At the point \(\left( {2,1} \right)\), the derivative \(y'\) is equal
\[y'\left( {2,1} \right) = \frac{{1 - {2^2}}}{{{1^2} - 2}} = 3.\]
The differential at this point is respectively written as
\[dy = y'dx = 3dx.\]
Example 20.
The function \(y\left( x \right)\) is defined by the implicit equation \[{x^2} - \sqrt y \,\ln y = 1.\] Find its differential at the point \(\left( {1,1} \right).\)
Solution.
We differentiate both sides of the equation with respect to \(x\) and find the derivative \(y':\)
\[\left( {{x^2} - \sqrt y \ln y} \right)' = 1',\;\; \Rightarrow 2x - \left( {\frac{1}{{2\sqrt y }} \cdot y' \cdot \ln y + \sqrt y \cdot \frac{1}{y} \cdot y'} \right) = 0,\;\; \Rightarrow 2x - y'\left( {\frac{{\ln y}}{{2\sqrt y }} + \frac{1}{{\sqrt y }}} \right) = 0,\;\; \Rightarrow 4x\sqrt y - y'\left( {\ln y + 2} \right) = 0,\;\; \Rightarrow y' = \frac{{4x\sqrt y }}{{\ln y + 2}}.\]
Calculate the value of the derivative at the given point \(\left( {1,1} \right):\)
\[y'\left( {1,1} \right) = \frac{{4 \cdot 1 \cdot \sqrt 1 }}{{\ln 1 + 2}} = 2.\]
The differential of the function at this point is written in the following form:
\[dy = y'dx = 2dx.\]
Example 21.
Suppose that a force \(F\) is determined by the law \[F = \frac{1}{{{r^2}}}.\] The distance \(r\) is measured as \(r = 2 \pm 0,2.\) Find the approximate value of the force \(F.\)
Solution.
Note that
\[F\left( {r = 2} \right) = \frac{1}{{{2^2}}} = \frac{1}{4} = 0,25.\]
We evaluate the differential \(dF\) at the point \(r = 2\) when \(dr = 0,2.\) As
\[dF = F^\prime\left( r \right)dr\]
and
\[F^\prime\left( r \right) = \left( {\frac{1}{{{r^2}}}} \right)^\prime = - \frac{2}{{{r^3}}},\]
then
\[dF = \left( { - \frac{2}{{{r^3}}}} \right)dr = - \frac{2}{{{2^3}}} \cdot 0,2 = - 0,05.\]
Similarly we can find that \(dF = 0,05\) for \(dr = -0,2.\) Therefore, the force \(F\) is estimated to be in the range
\[F = 0,25 \mp 0,05.\]
Example 22.
A body is moving through a liquid. The resistance force \(F\) depends on the speed of the body as \(F = k{v^2},\) where \(v\) is the speed and \(k\) is a constant. The speed \(v\) was measured as \(v = 10 \pm 1.\) Determine the approximate value of the resistance force \(F\) if \(k = 1.\)
Solution.
It is easy to see that
\[F\left( {v = 10} \right) = {10^2} = 100.\]
We estimate the approximate value of the force using the differential:
\[dF = F^\prime\left( v \right)dv.\]
Take the derivative:
\[F^\prime\left( v \right) = \left( {{v^2}} \right)^\prime = 2v,\]
so when \(v = 10,\) the derivative is equal to
\[F^\prime\left( {10} \right) = 2 \cdot 10 = 20.\]
Assuming \(dv = 1,\) we have
\[dF = F^\prime\left( v \right)dv = 20 \cdot 1 = 20.\]
Similarly, we can find the differential for \(dv = -1:\)
\[dF = F^\prime\left( v \right)dv = 20 \cdot \left( { - 1} \right) = - 20.\]
Thus, the resistance force takes on values in the interval
\[F = 100 \pm 20.\]
Example 23.
The function \(y\left( x \right)\) is defined by the parametric equations
\[
\left\{
\begin{aligned}
x &= t^2 + t + 1 \\
y &= t^3 - 2t
\end{aligned}
\right..
\]
Find the differential of the function at the point \(\left( {3, - 1} \right).\)
Solution.
We calculate the corresponding values of the parameter \(t\) from the equation \(3 = {t^2} + t + 1:\)
\[{t^2} + t - 2 = 0,\;\; \Rightarrow D = 9,\;\; \Rightarrow {t_{1,2}} = \frac{{ - 1 \pm 3}}{2}= 1, - 2.\]
Make sure that the value \(t = 1\) satisfies the condition \(y = -1.\)
\[{y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {{t^3} - 2t} \right)}^\prime }}}{{{{\left( {{t^2} + t + 1} \right)}^\prime }}} = \frac{{3{t^2} - 2}}{{2t + 1}}.\]
When \(t = 1,\) the derivative has the following value:
\[{y'_x}\left( {t = 1} \right) = \frac{{3 \cdot {1^2} - 2}}{{2 \cdot 1 + 1}} = \frac{1}{3}.\]
Thus, the differential of the function at the point \(\left( {3, - 1} \right)\) is expressed by the formula
\[dy = {y'_x}\,dx = \frac{{dx}}{3}.\]
Example 24.
The function \(y\left( x \right)\) is defined by the parametric equations
\[
\left\{
\begin{aligned}
x &= \left( {t + 2} \right){e^t} \\
y &= {e^{t + 1}}
\end{aligned}
\right..
\]
Find the differential of the function at the point \(\left( {2, e} \right).\)
Solution.
First we determine the value of the parameter \(t\), which corresponds to the point \(\left( {2, e} \right).\) It follows from the second equation that
\[e = {e^{t + 1}},\;\; \Rightarrow t + 1 = 1,\;\; \Rightarrow t = 0.\]
Check the value of \(x\) at \(t = 0:\)
\[x = \left( {0 + 2} \right){e^0} = 2.\]
Find the derivative of the parametric function:
\[{y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {{e^{t + 1}}} \right)}^\prime }}}{{{{\left( {\left( {t + 2} \right){e^t}} \right)}^\prime }}} = \frac{{{e^{t + 1}}}}{{1 \cdot {e^t} + \left( {t + 2} \right){e^t}}} = \frac{{\cancel{e^t}e}}{{\cancel{e^t}\left( {1 + t + 2} \right)}} = \frac{e}{{t + 3}}.\]
When \(t = 0,\) the derivative is respectively equal to
\[{y'_x}\left( {t = 0} \right) = \frac{e}{3}.\]
Consequently, the differential of the function at the point \(\left( {2, e} \right)\) has the form:
\[dy = {y'_x}\,dx = \frac{{e\,dx}}{3}.\]
Example 25.
Given the composite function \(y = \ln u,\;u = \cos x.\) Express the differential of \(y\) in an invariant form.
Solution.
We write the differential of the "outer" function:
\[dy = {y'_u}\,du = {\left( {\ln u} \right)^\prime }du = \frac{1}{u}du.\]
Similarly, we find the differential of the "inner" function:
\[du = {u'_x}\,dx = {\left( {\cos x} \right)^\prime }dx = - \sin x\,dx.\]
Substituting the expression for \(du\) in the previous formula, we obtain the differential \(dy\) in invariant form:
\[dy = \frac{1}{u}du = \frac{1}{u}\left( { - \sin x} \right)dx = - \frac{{\sin x}}{{\cos x}}dx = - \tan x\,dx.\]
