# Higher-Order Derivatives

## Solved Problems

### Example 9.

Find all derivatives of the function $y = {\frac{1}{x}}.$

Solution.

We find a few first derivatives:

$y' = {\left( {\frac{1}{x}} \right)^\prime } = - \frac{1}{{{x^2}}},$
$y^{\prime\prime} = \left( { - \frac{1}{{{x^2}}}} \right)^\prime = - 1 \cdot {\left( {{x^{ - 2}}} \right)^\prime } = - 1 \cdot \left( { - 2} \right) \cdot {x^{ - 3}} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2}}{{{x^3}}},$
$y^{\prime\prime\prime} = \left( {\frac{{{{\left( { - 1} \right)}^2} \cdot 2}}{{{x^3}}}} \right)^\prime = {\left( { - 1} \right)^2} \cdot 2 \cdot {\left( {{x^{ - 3}}} \right)^\prime } = {\left( { - 1} \right)^3} \cdot 2 \cdot 3 \cdot {x^{ - 4}} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 3}}{{{x^4}}},$
${y^{IV}} = \left( {\frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 3}}{{{x^4}}}} \right)^\prime = {\left( { - 1} \right)^3} \cdot 2 \cdot 3 \cdot {\left( {{x^{ - 4}}} \right)^\prime } = {\left( { - 1} \right)^4} \cdot 4! \cdot {x^{ - 5}} = \frac{{{{\left( { - 1} \right)}^4}4!}}{{{x^5}}}.$

This is enough to detect the general pattern:

${y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^n}n!}}{{{x^{n + 1}}}}.$

### Example 10.

Find all derivatives of the function $y = \frac{1}{{2x - 3}}.$

Solution.

We take the few first derivatives:

$y^\prime = \left( {\frac{1}{{2x - 3}}} \right)^\prime = \left( {{{\left( {2x - 3} \right)}^{ - 1}}} \right)^\prime = - {\left( {2x - 3} \right)^{ - 2}} \cdot \left( {2x - 3} \right)^\prime = - 1 \cdot {\left( {2x - 3} \right)^{ - 2}} \cdot 2 = - \frac{2}{{{{\left( {2x - 3} \right)}^2}}};$
$y^{\prime\prime} = \left( { - 1 \cdot {{\left( {2x - 3} \right)}^{ - 2}} \cdot 2} \right)^\prime = - 1 \cdot \left( { - 2} \right) \cdot {\left( {2x - 3} \right)^{ - 3}} \cdot {2^2} = 2!\,{\left( {2x - 3} \right)^{ - 3}}{2^2} = \frac{{2!\,{2^2}}}{{{{\left( {2x - 3} \right)}^3}}};$
$y^{\prime\prime\prime} = \left( {2!\,{{\left( {2x - 3} \right)}^{ - 3}}{2^2}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {\left( {2x - 3} \right)^{ - 4}} \cdot {2^3} = - 3!\,{\left( {2x - 3} \right)^{ - 4}}{2^3} = - \frac{{3!\,{2^3}}}{{{{\left( {2x - 3} \right)}^4}}};$
${y^{\left( 4 \right)}} = \left( { - 3!\,{{\left( {2x - 3} \right)}^{ - 4}}{2^3}} \right)^\prime = - 3! \cdot \left( { - 4} \right) \cdot {\left( {2x - 3} \right)^{ - 5}} \cdot {2^4} = 4!\,{\left( {2x - 3} \right)^{ - 5}}{2^4} = \frac{{4!\,{2^4}}}{{{{\left( {2x - 3} \right)}^5}}}.$

It is clear that the $$n$$th order derivative is written as

$y^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^n}n!\,{2^n}}}{{{{\left( {2x - 3} \right)}^{n + 1}}}}.$

### Example 11.

Find all derivatives of the function $y = \frac{1}{{1 - 5x}}.$

Solution.

Let's take the few first derivatives using the power rule and the chain rule:

$y^\prime = \left( {\frac{1}{{1 - 5x}}} \right)^\prime = \left( {{{\left( {1 - 5x} \right)}^{ - 1}}} \right)^\prime = - 1 \cdot {\left( {1 - 5x} \right)^{ - 2}} \cdot \left( {1 - 5x} \right)^\prime = - 1 \cdot {\left( {1 - 5x} \right)^{ - 2}} \cdot \left( { - 5} \right) = 1 \cdot {\left( {1 - 5x} \right)^{ - 2}} \cdot 5 = \frac{{1 \cdot 5}}{{{{\left( {1 - 5x} \right)}^2}}};$
$y^{\prime\prime} = \left( {1 \cdot {{\left( {1 - 5x} \right)}^{ - 2}} \cdot 5} \right)^\prime = 1 \cdot \left( { - 2} \right) \cdot {\left( {1 - 5x} \right)^{ - 3}} \cdot 5 \cdot \left( { - 5} \right) = 2! \cdot {\left( {1 - 5x} \right)^{ - 3}} \cdot {5^2} = \frac{{2!\,{5^2}}}{{{{\left( {1 - 5x} \right)}^3}}};$
$y^{\prime\prime\prime} = \left( {2! \cdot {{\left( {1 - 5x} \right)}^{ - 3}} \cdot {5^2}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {\left( {1 - 5x} \right)^{ - 4}} \cdot {5^2} \cdot \left( { - 5} \right) = 3! \cdot {\left( {1 - 5x} \right)^{ - 4}} \cdot {5^3} = \frac{{3!\,{5^3}}}{{{{\left( {1 - 5x} \right)}^4}}};$
${y^{\left( 4 \right)}} = \left( {3! \cdot {{\left( {1 - 5x} \right)}^{ - 4}} \cdot {5^3}} \right)^\prime = 3! \cdot \left( { - 4} \right) \cdot {\left( {1 - 5x} \right)^{ - 5}} \cdot {5^3} \cdot \left( { - 5} \right) = 4! \cdot {\left( {1 - 5x} \right)^{ - 5}} \cdot {5^4} = \frac{{4!\,{5^4}}}{{{{\left( {1 - 5x} \right)}^5}}}.$

So, the derivative of the $$n$$th order is given by

${y^{\left( n \right)}} = \frac{{n!\,{5^n}}}{{{{\left( {1 - 5x} \right)}^{n + 1}}}}.$

### Example 12.

Find the $$n$$th order derivative of the function $y = {3^{2x + 1}}.$

Solution.

We calculate successively several derivatives, starting from the first one.

$y' = {\left( {{3^{2x + 1}}} \right)^\prime } = {3^{2x + 1}} \cdot \ln 3 \cdot {\left( {2x + 1} \right)^\prime } = {3^{2x + 1}} \cdot 2\ln 3,$
$y^{\prime\prime} = {\left( {y'} \right)^\prime } = {\left( {{3^{2x + 1}} \cdot 2\ln 3} \right)^\prime } = {\left( {{3^{2x + 1}}} \right)^\prime } \cdot 2\ln 3 = {3^{2x + 1}} \cdot {2^2}{\ln ^2}3,$
$y^{\prime\prime\prime} = {\left( {y^{\prime\prime}} \right)^\prime } = {\left( {{3^{2x + 1}} \cdot {2^2}{{\ln }^2}3} \right)^\prime } = {\left( {{3^{2x + 1}}} \right)^\prime } \cdot {2^2}{\ln ^2}3 = {3^{2x + 1}} \cdot {2^3}{\ln ^3}3.$

It follows from here that the $$n$$th order derivative is given by

${y^{\left( n \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( n \right)}} = {3^{2x + 1}} \cdot {2^n}{\ln ^n}3.$

A rigorous proof can be carried out by induction. Clearly that this formula is valid for $$n = 1$$. Suppose that it is true for $$n = k:$$

${y^{\left( k \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( k \right)}} = {3^{2x + 1}} \cdot {2^k}\,{\ln ^k}3.$

Then for $$n = k + 1,$$ we have

${y^{\left( {k + 1} \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( {k + 1} \right)}} = {\left[ {{{\left( {{3^{2x + 1}}} \right)}^{\left( k \right)}}} \right]^\prime } = {\left( {{3^{2x + 1}} \cdot {2^k}\,{{\ln }^k}3} \right)^\prime } = {\left( {{3^{2x + 1}}} \right)^\prime } \cdot {2^k}\,{\ln ^k}3 = {3^{2x + 1}} \cdot {2^{k + 1}}{\ln ^{k + 1}}3,$

that is, this formula holds for $$n = k + 1.$$ Consequently, it is true for any natural number $$n.$$

### Example 13.

Find the $$n$$th derivative of the function $y = x{e^x}$ at $$x = 0.$$

Solution.

We differentiate successively using the product rule:

$y^\prime = \left( {x{e^x}} \right)^\prime = 1 \cdot {e^x} + x \cdot {e^x} = \left( {x + 1} \right){e^x};$
$y^{\prime\prime} = \left( {\left( {x + 1} \right){e^x}} \right)^\prime = 1 \cdot {e^x} + \left( {x + 1} \right) \cdot {e^x} = \left( {x + 2} \right){e^x};$
$y^{\prime\prime\prime} = \left( {\left( {x + 2} \right){e^x}} \right)^\prime = 1 \cdot {e^x} + \left( {x + 2} \right) \cdot {e^x} = \left( {x + 3} \right){e^x}.$

Hence

$y^{\left( n \right)} = \left( {x + n} \right){e^x},$

where $$n$$ is a whole number. Then

${y^{\left( n \right)}}\left( 0 \right) = \left( {0 + n} \right){e^0} = n \cdot 1 = n.$

### Example 14.

Find the $$n$$th order derivative of the power function $y = {x^m}$ where $$m$$ is a real number.

Solution.

We calculate several first derivatives:

$y' = {\left( {{x^m}} \right)^\prime } = m{x^{m - 1}},$
$y^{\prime\prime} = \left( {y'} \right)^\prime = \left( {m{x^{m - 1}}} \right)^\prime = m\left( {m - 1} \right){x^{m - 2}},$
$y^{\prime\prime\prime} = \left( {y^{\prime\prime}} \right)^\prime = \left[ {m\left( {m - 1} \right){x^{m - 2}}} \right]^\prime = \left[ {m\left( {m - 1} \right)\left( {m - 2} \right)} \right]{x^{m - 3}}.$

Hence, it is easy to establish a general expression for the $$n$$th order derivative:

$y^{\left( n \right)} = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n + 1} \right){x^{m - n}}.$

We prove this by induction. Obviously, this formula is valid for $$n = 1$$. Assuming that it is true for the degree $$n$$, we differentiate it again and find the derivative of $$\left( {n + 1} \right)$$th order:

$y^{\left( {n + 1} \right)} = {\left[ {{y^{\left( n \right)}}} \right]^\prime } = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n + 1} \right){\left( {{x^{m - n}}} \right)^\prime } = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n} \right){x^{m - \left( {n + 1} \right)}}.$

As you can see, the derivative of $$\left( {n + 1} \right)$$th order is expressed by the same formula as the $$n$$th order derivative (only the number $$n$$ is replaced by $$n + 1$$). Consequently, the resulting expression is valid for any positive integer value of $$n$$ ($$n$$ is the order of the derivative).

Note that the exponent $$m$$, generally speaking, is a real number. If we consider only the natural values of $$m,$$ then the formula for the derivative can be written in a more compact form:

$y^{\left( n \right)} = \left( {{x^m}} \right)^{\left( n \right)} = \frac{{m!}}{{\left( {m - n} \right)!}}{x^{m - n}},$

where $$n \le m$$. All other derivatives of order $$n \gt m$$ are equal to zero.

### Example 15.

Find the $$n$$th order derivative of the square root $y = \sqrt x .$

Solution.

We use the result of Example $$14$$, where the $$n$$th order derivative of the power function with an arbitrary real exponent $$m$$ is derived. In this case we have

$y = \sqrt x = {x^{\frac{1}{2}}}\;\;\left( {m = \frac{1}{2}} \right).$

Then the $$n$$th order derivative is written as

$y^{\left( n \right)} = \left( {{x^{\frac{1}{2}}}} \right)^{\left( n \right)} = \frac{1}{2} \cdot \left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right) \cdots \left( {\frac{1}{2} - n + 1} \right){x^{\frac{1}{2} - n}} = \frac{1}{2} \cdot \left( { - \frac{1}{2}} \right) \cdot \left( { - \frac{3}{2}} \right) \cdot \left( { - \frac{5}{2}} \right) \cdots \left[ { - \left( {n - \frac{3}{2}} \right)} \right]\frac{{{x^{\frac{1}{2}}}}}{{{x^n}}} = \frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}} \left[ {\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2} \cdots \frac{{2n - 3}}{2}} \right]\frac{{\sqrt x }}{{{x^n}}} = \frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}} \cdot \frac{1}{{{2^{n - 1}}}} \cdot \left[ {1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right)} \right]\frac{{\sqrt x }}{{{x^n}}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left[ {1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right)} \right]\sqrt x }}{{{x^n}}}.$

For $$n = 1$$, the derivative is

$y' = \frac{{{{\left( { - 1} \right)}^0} \cdot 1 \cdot \sqrt x }}{{{{\left( {2x} \right)}^1}}} = \frac{{\sqrt x }}{{2x}} = \frac{1}{{2\sqrt x }}.$

Provided $$n \ge 2$$, the product of odd numbers in square brackets can be written in terms of the double factorial:

$1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right) = \left( {2n - 3} \right)!!$

Hence for $$n \ge 2$$, the $$n$$th order derivative is expressed by the general formula

${y^{\left( n \right)}} = \frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}} \cdot \frac{1}{{{2^{n - 1}}}} \cdot \left( {2n - 3} \right)!!\frac{{\sqrt x }}{{{x^n}}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {2n - 3} \right)!!\sqrt x }}{{{{\left( {2x} \right)}^n}}}.$

In particular,

$y^{\prime\prime} = \frac{{{{\left( { - 1} \right)}^1}1!!\sqrt x }}{{{{\left( {2x} \right)}^2}}} = - \frac{{\sqrt x }}{{4{x^2}}} = - \frac{1}{{4\sqrt {{x^3}} }},$
$y^{\prime\prime\prime} = \frac{{{{\left( { - 1} \right)}^2}3!!\sqrt x }}{{{{\left( {2x} \right)}^3}}} = \frac{{3\sqrt x }}{{8{x^3}}} = \frac{3}{{8\sqrt {{x^5}} }}.$

### Example 16.

Find the $$n$$th order derivative of the cube root $y = \sqrt[3]{x}.$

Solution.

The first derivative of the cube root is given by

$y' = \left( {\sqrt[3]{x}} \right)^\prime = \left( {{x^{\frac{1}{3}}}} \right)^\prime = \frac{1}{3}{x^{\frac{1}{3} - 1}} = \frac{1}{3}{x^{ - \frac{2}{3}}} = \frac{1}{{3\sqrt[3]{{{x^2}}}}}.$

Next, we use the general formula for the $$n$$th order derivative of the power function $$y = {x^m}$$ (Example $$14$$):

$\left( {{x^m}} \right)^{\left( n \right)} = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n + 1} \right){x^{m - n}}.$

In our case $$m = {\frac{1}{3}}.$$ Consequently, the derivative is as follows:

$y^{\left( n \right)} = \left( {\sqrt[3]{x}} \right)^{\left( n \right)} = \left( {{x^{\frac{1}{3}}}} \right)^{\left( n \right)} = \frac{1}{3}\left( {\frac{1}{3} - 1} \right) \left( {\frac{1}{3} - 2} \right)\left( {\frac{1}{3} - 3} \right) \cdots \left( {\frac{1}{3} - n + 1} \right){x^{\frac{1}{3} - n}} = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) \cdot \left( { - \frac{5}{3}} \right) \cdot \left( { - \frac{8}{3}} \right) \cdots \left[ { - \left( {n - \frac{4}{3}} \right)} \right]\frac{{{x^{\frac{1}{3}}}}}{{{x^n}}} = \frac{1}{3} \cdot {\left( { - 1} \right)^{n - 1}} \left[ {\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdots \frac{{3n - 4}}{3}} \right]\frac{{\sqrt[3]{x}}}{{{x^n}}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left[ {2 \cdot 5 \cdot 8 \cdots \left( {3n - 4} \right)} \right]\sqrt[3]{x}}}{{{{\left( {3x} \right)}^n}}},$

where $$n \ge 2.$$

Specifically, the second and third derivatives of the cube root are expressed by the formulas

$y^{\prime\prime} = \frac{{{{\left( { - 1} \right)}^1} \cdot 2 \cdot \sqrt[3]{x}}}{{{{\left( {3x} \right)}^2}}} = - \frac{{2\sqrt[3]{x}}}{{9{x^2}}} = - \frac{2}{{9\sqrt[3]{{{x^5}}}}},$
$y^{\prime\prime\prime} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 5 \cdot \sqrt[3]{x}}}{{{{\left( {3x} \right)}^3}}} = \frac{{10\sqrt[3]{x}}}{{27{x^3}}} = \frac{{10}}{{27\sqrt[3]{{{x^8}}}}}.$

### Example 17.

Find the $$3$$rd derivative of the function $$y = f\left( x \right)$$ given by the equation ${y^2} = 2x.$

Solution.

Differentiate both sides with respect to $$x$$ to calculate the first derivative $$y_x^\prime:$$

$\left( {{y^2}} \right)^\prime = \left( {2x} \right)^\prime,\;\; \Rightarrow 2yy^\prime = 2,\;\; \Rightarrow yy^\prime = 1,\;\; \Rightarrow y^\prime = y_x^\prime = \frac{1}{y}.$

The second derivative $$y_{xx}^{\prime\prime}$$ is given by

$\left( {yy^\prime} \right)^\prime = 1^\prime,\;\; \Rightarrow y^\prime y^\prime + yy^{\prime\prime} = 0,\;\; \Rightarrow yy^{\prime\prime} = - \left( {y^\prime} \right)^2,\;\; \Rightarrow y^{\prime\prime} = y_{xx}^{\prime\prime} = - \frac{{{{\left( {y^\prime} \right)}^2}}}{y} = - \frac{{{{\left( {\frac{1}{y}} \right)}^2}}}{y} = - \frac{1}{{{y^3}}}.$

Differentiating once more with respect to $$x,$$ we find the $$3$$rd derivative $$y_{xxx}^{\prime\prime\prime}:$$

$\left( {{{\left( {y^\prime} \right)}^2} + yy^{\prime\prime}} \right)^\prime = 0^\prime,\;\; \Rightarrow 2y^\prime y^{\prime\prime} + y^\prime y^{\prime\prime} + yy^{\prime\prime\prime} = 0,\;\; \Rightarrow 3y^\prime y^{\prime\prime} + yy^{\prime\prime\prime} = 0,\;\; \Rightarrow y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = - \frac{{3y^\prime y^{\prime\prime}}}{y} = - \frac{{3 \cdot \frac{1}{y} \cdot \left( { - \frac{1}{{{y^3}}}} \right)}}{y} = \frac{3}{{{y^5}}}.$

### Example 18.

Given the equation of an ellipse in parametric form: $x = a\cos t,\;y = b\sin t,$ where $$a$$, $$b$$ are semi-axes of the ellipse, $$t$$ is a parameter. Find the first, second and third derivatives of the function $$y$$ with respect to $$x.$$

Solution.

Differentiating the given parametric function successively, we obtain:

$y^\prime = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {b\sin t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} = \frac{{b\cos t}}{{ - a\sin t}} = - \frac{b}{a}\cot t,$
$y^{\prime\prime} = y^{\prime\prime}_{xx} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( { - \frac{b}{a}\cot t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} = \frac{{\left( { - \frac{b}{a}} \right)\left( { - \frac{1}{{{{\sin }^2}t}}} \right)}}{{\left( { - a\sin t} \right)}} = - \frac{b}{{{a^2}}}\frac{1}{{{{\sin }^3}t}} = - \frac{b}{{{a^2}}}{\csc ^3}t,$
$y^{\prime\prime\prime} = y^{\prime\prime\prime}_{xxx} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}'_t} }}{{{x'_t}}} = \frac{{{{\left( { - \frac{b}{{{a^2}}}{{\csc }^3}t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} = \frac{{\left( { - \frac{b}{{{a^2}}}} \right) \cdot 3{{\csc }^2}t \cdot {{\left( {\csc t} \right)}^\prime }}}{{\left( { - a\sin t} \right)}} = - \frac{{3b}}{{{a^3}}} \cdot \frac{{{{\csc }^2}t \cdot \left( { - \cot t} \right) \cdot \csc t}}{{\left( { - \sin t} \right)}} = - \frac{{3b}}{{{a^3}}} \cdot \frac{{{{\csc }^3}t\cot t}}{{\sin t}} = - \frac{{3b}}{{{a^3}}}{\csc ^4}t\cot t.$

### Example 19.

Find the $$3$$rd derivative of the function given by the parametric equations $x = 1 + {t^2},\; y = t - {t^3}$ at $$t = 1.$$

Take the first derivative:

$y^\prime = y_x^\prime = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\left( {t - {t^3}} \right)^\prime}}{{\left( {1 + {t^2}} \right)^\prime}} = \frac{{1 - 3{t^2}}}{{2t}} = \frac{1}{{2t}} - \frac{{3t}}{2}.$

Continue differentiating:

$y^{\prime\prime} = y_{xx}^{\prime\prime} = \frac{{\left( {y_x^\prime} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( {\frac{1}{{2t}} - \frac{{3t}}{2}} \right)^\prime}}{{\left( {1 + {t^2}} \right)^\prime}} = \frac{{\frac{1}{2} \cdot \left( { - {t^{ - 2}}} \right) - \frac{3}{2}}}{{2t}} = - \frac{{\frac{1}{{2{t^2}}} + \frac{3}{2}}}{{2t}} = - \frac{1}{{4{t^3}}} - \frac{3}{{4t}}.$

Similarly we calculate the third derivative $$y_{xxx}^{\prime\prime\prime}:$$

$y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = \frac{{\left( {y_{xx}^{\prime\prime}} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( { - \frac{1}{{4{t^3}}} - \frac{3}{{4t}}} \right)^\prime}}{{\left( {1 + {t^2}} \right)^\prime}} = \frac{{ - \frac{1}{4} \cdot \left( { - 3{t^{ - 4}}} \right) - \frac{3}{4} \cdot \left( { - {t^{ - 2}}} \right)}}{{2t}} = \frac{{\frac{3}{{4{t^4}}} + \frac{3}{{4{t^2}}}}}{{2t}} = \frac{3}{{8{t^5}}} + \frac{3}{{8{t^3}}}.$

At the point where $$t = 1,$$ the third derivative is equal to

$y_{xxx}^{\prime\prime\prime}\left( {t = 1} \right) = \frac{3}{{8 \cdot {1^5}}} + \frac{3}{{8 \cdot {1^3}}} = \frac{3}{8} + \frac{3}{8} = \frac{3}{4}.$

### Example 20.

Find the $$3$$rd derivative of the function given by the parametric equations $x = 1 + \sin t,\;y = t - \cos t$ at $$t = 0.$$

Solution.

Compute the first derivative $$y_x^\prime:$$

$y^\prime = y_x^\prime = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\left( {t - \cos t} \right)^\prime}}{{\left( {1 + \sin t} \right)^\prime}} = \frac{{1 + \sin t}}{{\cos t}}.$

Similarly we determine the second derivative $$y_{xx}^{\prime\prime}$$ of the parametric function:

$y^{\prime\prime} = y_{xx}^{\prime\prime} = \frac{{\left( {y_x^\prime} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( {\frac{{1 + \sin t}}{{\cos t}}} \right)^\prime}}{{\left( {1 + \sin t} \right)^\prime}} = \frac{{\frac{{\cos t \cdot \cos t - \left( {1 + \sin t} \right) \cdot \left( { - \sin t} \right)}}{{{{\cos }^2}t}}}}{{\cos t}} = \frac{{{{\cos }^2}t + \sin t + {{\sin }^2}t}}{{{{\cos }^3}t}} = \frac{{1 + \sin t}}{{{{\cos }^3}t}}.$

The third derivative $$y_{xxx}^{\prime\prime\prime}$$ has the form

$y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = \frac{{\left( {y_{xx}^{\prime\prime}} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( {\frac{{1 + \sin t}}{{{{\cos }^3}t}}} \right)^\prime}}{{\left( {1 + \sin t} \right)^\prime}} = \frac{{{{\cos }^4}t + 3\,{{\cos }^2}t\sin t + 3\,{{\cos }^2}t\,{{\sin }^2}t}}{{{{\cos }^7}t}} = \frac{{{{\cos }^2}t\left( {{{\cos }^2}t + 3\sin t + 3\,{{\sin }^2}t} \right)}}{{{{\cos }^7}t}} = \frac{{{{\cos }^2}t + 3\sin t + 3\,{{\sin }^2}t}}{{{{\cos }^5}t}}.$

So, at the point where $$t = 0,$$ the third derivative is equal to

$y_{xxx}^{\prime\prime\prime}\left( {t = 0} \right) = \frac{{{{\cos }^2}0 + 3\sin 0 + 3\,{{\sin }^2}0}}{{{{\cos }^5}0}} = \frac{{1 + 0 + 0}}{1} = 1.$

### Example 21.

Find the third derivative of the function given by the equation ${x^2} + 3xy + {y^2} = 1.$

Solution.

Differentiate one time with respect to $$x:$$

$\left( {{x^2} + 3xy + {y^2}} \right)^\prime = {1^\prime },\;\; \Rightarrow 2x + 3\left( {x'y + xy'} \right) + 2yy' = 0,\;\; \Rightarrow 2x + 3y + 3xy' + 2yy' = 0,\;\; \Rightarrow 2x + 3y + \left( {3x + 2y} \right)y' = 0,\; \Rightarrow y' = - \frac{{2x + 3y}}{{3x + 2y}}.$

Now we differentiate the last expression again considering $$y$$ as a composite function:

$y^{\prime\prime} = {\left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)^\prime } = \frac{{5y - 5xy'}}{{{{\left( {3x + 2y} \right)}^2}}}.$

Substitute the explicit expression for the first derivative $$y':$$

$y^{\prime\prime} = \frac{{5y - 5xy'}}{{{{\left( {3x + 2y} \right)}^2}}} = \frac{{5y - 5x\left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)}}{{{{\left( {3x + 2y} \right)}^2}}} = \frac{{5y\left( {3x + 2y} \right) + 5x\left( {2x + 3y} \right)}}{{{{\left( {3x + 2y} \right)}^3}}} = \frac{{15xy + 10{y^2} + 10{x^2} + 15xy}}{{{{\left( {3x + 2y} \right)}^3}}} = \frac{{10\left( {{x^2} + 3xy + {y^2}} \right)}}{{{{\left( {3x + 2y} \right)}^3}}}.$

Since $${x^2} + 3xy + {y^2} = 1,$$ we obtain the following expression for the second derivative $$y^{\prime\prime}:$$

$y^{\prime\prime} = \frac{{10}}{{{{\left( {3x + 2y} \right)}^3}}}.$

Similarly, differentiating once more, we get the third derivative:

$y^{\prime\prime\prime} = {\left[ {\frac{{10}}{{{{\left( {3x + 2y} \right)}^3}}}} \right]^\prime } = - \frac{{30\left( {3 + 2y'} \right)}}{{{{\left( {3x + 2y} \right)}^4}}} = - \frac{{90 + 60y'}}{{{{\left( {3x + 2y} \right)}^4}}}.$

We substitute again the first derivative and find:

$y^{\prime\prime\prime} = - \frac{{90 + 60y'}}{{{{\left( {3x + 2y} \right)}^4}}} = - \frac{{90 + 60 \cdot \left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)}}{{{{\left( {3x + 2y} \right)}^4}}} = - \frac{{90\left( {3x + 2y} \right) - 60\left( {2x + 3y} \right)}}{{{{\left( {3x + 2y} \right)}^5}}} = - \frac{{150}}{{{{\left( {3x + 2y} \right)}^5}}}.$

### Example 22.

Find the $$3$$rd derivative of the function $$y = f\left( x \right)$$ given by the equation ${x^2} - {y^2} = 9.$

Solution.

We differentiate both sides of the equation with respect to $$x$$ keeping in mind that $$y$$ is a function of $$x:$$

$\left( {{x^2} - {y^2}} \right)^\prime = 9^\prime,\;\; \Rightarrow 2x - 2yy^\prime = 0, \Rightarrow x - yy^\prime = 0,\; \Rightarrow y^\prime = y_x^\prime = \frac{x}{y}.$

Continue differentiating to obtain $$y_{xx}^{\prime\prime}:$$

$x - yy^\prime = 0,\;\; \Rightarrow \left( {x - yy^\prime} \right)^\prime = 0,\;\; \Rightarrow 1 - y^\prime y^\prime - yy^{\prime\prime} = 0,\;\; \Rightarrow yy^{\prime\prime} = 1 - {\left( {y^\prime} \right)^2},\;\; \Rightarrow y^{\prime\prime} = y_{xx}^{\prime\prime} = \frac{{1 - {{\left( {y^\prime} \right)}^2}}}{y} = \frac{{1 - {{\left( {\frac{x}{y}} \right)}^2}}}{y} = \frac{{1 - \frac{{{x^2}}}{{{y^2}}}}}{y} = \frac{{{y^2} - {x^2}}}{{{y^3}}} = - \frac{{{x^2} - {y^2}}}{{{y^3}}} = - \frac{9}{{{y^3}}}.$

Similarly we find the third derivative:

$yy^{\prime\prime} = 1 - {\left( {y^\prime} \right)^2},\;\; \Rightarrow \left( {yy^{\prime\prime}} \right)^\prime = \left( {1 - {{\left( {y^\prime} \right)}^2}} \right)^\prime,\;\; \Rightarrow y^\prime y^{\prime\prime} + yy^{\prime\prime\prime} = - 2y^\prime y^{\prime\prime},\;\; \Rightarrow yy^{\prime\prime\prime} = - 3y^\prime y^{\prime\prime},\;\; \Rightarrow y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = - \frac{{3y^\prime y^{\prime\prime}}}{y} = - \frac{{3 \cdot \frac{x}{y} \cdot \left( { - \frac{9}{{{y^3}}}} \right)}}{y} = \frac{{\frac{{27x}}{{{y^4}}}}}{y} = \frac{{27x}}{{{y^5}}}.$