# 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:

$dy = \frac{1}{u}du = \frac{1}{u}\left( { - \sin x} \right)dx = - \frac{{\sin x}}{{\cos x}}dx = - \tan x\,dx.$