Calculation of higher-order derivatives.

The second-order derivative of a function $y=f(x)$ is called the derivative of its first derivative, i.e., $$y^{″}(x)=(y^{\prime }(x){)}^{\prime }.$$

The $n$-th order derivative (or $n$-th derivative) is defined as the derivative of the $(n-1)$-th order derivative, i.e., $$y^{(n)}(x)={(y^{(n-1)}(x))}^{\prime },n=2,3,...$$ For the $n$-th order derivative, the notation $\frac{d^{n}y}{d{x}^n}$ is used.

Let $u(x)$ and $v(x)$ have derivatives up to the $n$-th order inclusive. Then, for the $n$-th order derivative of their product $u(x)v(x)$, the Leibniz formula applies:a $$(uv{)}^{(n)}=u^{n}v+n{u}^{(n-1)}{v}^{\prime }+\frac{n(n-1)}{1\cdot 2}{u}^{(n-1)}{v}^{″}+...+u{v}^{(n)}=\sum _{k=0}^n{C}_n^k{u}^{(n-k)}{v}^{(k)},$$ where $u^{(0)}=u,,,v^{(0)}=v$ and $C_n^k=\frac{n(n-1)...(n-k+1)}{1\cdot 2\cdot ...\cdot k}=\frac{k!}{(n-k)!}$ are binomial coefficients (by definition, $0!=1$).

Examples.

Find the second-order derivatives of the following functions:

1. $y={\cos }^{2}x.$

Solution.

$y^{\prime }=({\cos }^{2}x{)}^{\prime }=2\cos x(\cos x{)}^{\prime }=-2\cos x\sin x=-\sin 2x;$

$y^{″}=(-\sin 2x{)}^{\prime }=-\cos 2x(2x{)}^{\prime }=-2\cos 2x.$

Answer: $-2\cos 2x.$

2. $y=e^{-x^2}.$

Solution.

$y^{\prime }=(e^{-x^2}{)}^{\prime }=e^{-x^2}(-x^2{)}^{\prime }=e^{-x^2}(-2x)=-2x{e}^{-x^2}.$

$$y^{″}=(-2x{e}^{-x^2}{)}^{″}=(-2x{)}^{\prime }{e}^{-x^2}-2x(e^{-x^2}{)}^{\prime }=-2e^{-x^2}-2x{e}^{-x^2}(-2x)=$$ $$=-2e^{-x^2}+4x^2{e}^{-x^2}.$$

Answer: $y^{″}=-2e^{-x^2}+4x^2{e}^{-x^2}.$

3. Find $y^{‴}(2),$ если $y=\ln (x-1).$

Solution.

$y^{\prime }=(\ln (x-1){)}^{\prime }=\frac{1}{x-1}(x-1{)}^{\prime }=\frac{1}{x-1};$

$y^{″}={\left(\frac{1}{x-1}\right)}^{\prime }=((x-1{)}^{-1}{)}^{\prime }=-1(x-1{)}^{-2}(x-1{)}^{\prime }=-(x-1{)}^{-2};$

$y^{‴}=(-(x-1{)}^{-2}{)}^{\prime }=-(-2)(x-1{)}^{-3}=2\frac{1}{(x-1{)}^3};$

$y^{‴}(2)=2\frac{1}{(2-1{)}^3}=2.$

Answer: 2.

Let $u(x)$ and $v(x)$ be twice differentiable functions. Find $y^{\prime },,,y^{″}$ if:

4. $y=\sqrt{u^2+v^2}.$

Solution.

$$y^{\prime }=(\sqrt{u^2+v^2}{)}^{\prime }=((u^2+v^2{)}^{1/2}{)}^{\prime }=\frac{1}{2}(u^2+v^2{)}^{-1/2}(u^2+v^2{)}^{\prime }=$$ $$=\frac{1}{2}(u^2+v^2{)}^{-1/2}(2u{u}^{\prime }+2v{v}^{\prime })=\frac{u{u}^{\prime }+v{v}^{\prime }}{\sqrt{u^2+v^2}};$$

$$y^{″}={((u^2+v^2{)}^{-1/2}(u{u}^{\prime }+v{v}^{\prime }))}^{\prime }=$$ $$=(-\frac{1}{2})(u^2+v^2{)}^{-1/2-1}(u^2+v^2{)}^{\prime }(u{u}^{\prime }+v{v}^{\prime })+$$ $$+(u^2+v^2{)}^{-1/2}(u^{\prime }{u}^{\prime }+u{u}^{″}+v^{\prime }{v}^{\prime }+v{v}^{″})=$$ $$=-\frac{1}{2}(u^2+v^2{)}^{-3/2}(u{u}^{\prime }+v{v}^{\prime })(2u{u}^{\prime }+2v{v}^{\prime })+$$ $$+(u^2+v^2{)}^{-1/2}(u^{\prime 2}+u{u}^{″}+v^{\prime 2}+v{v}^{″})=$$ $$-(u^2+v^2{)}^{-3/2}(u{u}^{\prime }+v{v}^{\prime }{)}^2+(u^2+v^2{)}^{-1/2}(u^{\prime 2}+u{u}^{″}+v^{\prime 2}+v{v}^{″})=$$ $$=\frac{-(u{u}^{\prime }+v{v}^{\prime })+(u^2+v^2)(u^{\prime 2}+u{u}^{″}+v^{\prime 2}+v{v}^{″})}{(u^2+v^2{)}^{3/2}}$$

Answer: $\frac{-(u{u}^{\prime }+v{v}^{\prime })+(u^2+v^2)(u^{\prime 2}+u{u}^{″}+v^{\prime 2}+v{v}^{″})}{(u^2+v^2{)}^{3/2}}.$

Find the formula for the $n$-th derivative of the given functions:

5. $y=x^m,m\in N.$

Solution.

$y^{\prime }=(x^m{)}^{\prime }=m{x}^{m-1};$

$y^{″}=(m{x}^{m-1}{)}^{\prime }=m(m-1)x^{m-2};$

$y^{‴}=(m(m-1)x^{m-2}{)}^{\prime }=m(m-1)(m-2)x^{m-3};$

...

$y^{(n)}=m(m-1)(m-2)...(m-n+1)x^{m-n}=\frac{m!}{(m-n+1)!}{x}^{m-n},n\le m$

$y^{(n)}=m(m-1)(m-2)...(m-m)...(m-n+1)x^{m-n}=0,n>m.$

Answer: $y^{(n)}=\frac{m!}{(m-n+1)!}{x}^{m-n},n\le m;$ $y^{(n)}=0,n>m.$

6. $y=\sin x.$

Solution.

$y^{\prime }=(\sin x{)}^{\prime }=\cos x=\sin (x+\frac{\pi }{2});$

$y^{″}=(\cos x{)}^{\prime }=-\sin x=\sin (x+\pi );$

$y^{‴}=(-\sin x{)}^{\prime }=-\cos x=\sin (x+\frac{3\pi }{2});$

$y^{(4)}=(-cosx{)}^{\prime }=\sin x=\sin (x+2\pi )=\sin (x+\frac{4\pi }{2});$

...

$y^{(n)}=\sin (x+\frac{\pi }{2}n).$

Answer: $y^{(n)}=\sin (x+\frac{\pi }{2}n).$

7. $y=\frac{1+x}{1-x}.$

Solution.

$$y^{\prime }={\left(\frac{1+x}{1-x}\right)}^{\prime }=\frac{(1+x{)}^{\prime }(1-x)-(1-x{)}^{\prime }(1+x)}{(1-x{)}^2}=$$ $$=\frac{1-x+1+x}{(1-x{)}^2}=\frac{2}{(1-x{)}^2}.$$

$$y^{″}={\left(\frac{2}{(1-x{)}^2}\right)}^{\prime }=2{((1-x{)}^{-2}))}^{\prime }=2\cdot (-2)(1-x{)}^{-2-1}(1-x{)}^{\prime }=$$ $$=2\cdot 2(1-x{)}^{-3}.$$

$$y^{‴}={(2\cdot 2(1-x{)}^{-3})}^{\prime }=2\cdot 2\cdot (-3)(1-x{)}^{-4}(-1)=2\cdot 2\cdot 3(1-x{)}^{-4}.$$

$$...$$

$$y^{(n)}=2n!(1-x{)}^{-n-1}.$$

Answer: $y^{(n)}=2n!(1-x{)}^{-n-1}.$

8. By decomposing into a linear combination of simpler functions, find the derivative $y^{(50)}$ of the function $y=\frac{1}{x^2-3x+2}.$

Solution.

Let's decompose the fraction $\frac{1}{x^2-3x+2}$ into elementary fractions:

$x^2-3x+2=0$

$D=9-8=1$

$x_1=\frac{3+1}{2}=2;$ $x_2=\frac{3-1}{2}=1.$

Thus, $x^2-3x+2=(x-1)(x-2).$

Hence, $\frac{1}{x^2-3x+2}=\frac{A}{x-1}+\frac{B}{x-2}=\frac{Ax-2A+Bx-B}{(x-1)(x-2)}.$

Therefore, $1=Ax-2A+Bx-B.$

We equate the coefficients of like terms:

$$\{\begin{array}{l}1=-2A-B\\ 0=A+B\end{array}\Rightarrow \{\begin{array}{l}1=2B-B\\ B=-A\end{array}\Rightarrow \{\begin{array}{l}B=1\\ A=-1\end{array}$$

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

$y^{″}={\left(\frac{1}{x^2-3x+2}\right)}^{″}={(-\frac{1}{x-1}+\frac{1}{x-2})}^{″}=$ $=-((x-1{)}^{-1}{)}^{″}+((x-2{)}^{-1}{)}^{″}=((x-1{)}^{-2}-(x-2{)}^{-2}{)}^{\prime }=$ $=-2(x-1{)}^{-3}+2(x-2{)}^{-3}.$

$y^{‴}={\left(\frac{1}{x^2-3x+2}\right)}^{‴}={(-\frac{1}{x-1}+\frac{1}{x-2})}^{‴}=$ $=-((x-1{)}^{-1}{)}^{‴}+((x-2{)}^{-1}{)}^{‴}=(-2(x-1{)}^{-3}+2(x-2{)}^{-3}{)}^{\prime }=$ $=2\cdot 3(x-1{)}^{-4}-2\cdot 3(x-2{)}^{-4}.$

......

$y^{50}={\left(\frac{1}{x^2-3x+2}\right)}^{(50)}={(-\frac{1}{x-1}+\frac{1}{x-2})}^{(50)}=$ $=-((x-1{)}^{-1}{)}^{(50)}+((x-2{)}^{-1}{)}^{(50)}=(-2(x-1{)}^{-3}+2(x-2{)}^{-3}{)}^{\prime }=$ $=-50!(x-1{)}^{-51}+50!(x-2{)}^{-51}=50!(-\frac{1}{(x-1{)}^{51}}+\frac{1}{(x-2{)}^{51}}).$

Answer: $y^{(50)}=50!(-\frac{1}{(x-1{)}^{51}}+\frac{1}{(x-2{)}^{51}})$.

(The answer did not match the one in the book -- their solution is $y^{50}=\frac{(x-1{)}^{50}-(x-2{)}^{50}}{(x^2-3x+2{)}^{51}}.$)

Applying the Leibniz formula, find the specified derivatives of the given functions:

9. $y=(x^2+x+1)\sin x$, find $y^{(15)}.$

Solution.

Using the Leibniz formula, we get:

$$y^{(15)}={((x^2+x+1)\sin x)}^{(15)}=(\sin x{)}^{(15)}(x^2+x+1)+$$ $$+15(\sin x{)}^{(14)}(x^2+x+1{)}^{\prime }+\frac{15\cdot 14}{1\cdot 2}(\sin x{)}^{(13)}(x^2+x+1{)}^{″}+$$ $$+\frac{15\cdot 14\cdot 13}{1\cot 2\cdot 3}(\sin x{)}^{(12)}(x^2+x+1{)}^{‴}+...+\sin x(x^2+x+1{)}^{(15)}$$

$$(x^2+x+1{)}^{\prime }=2x+1;$$

$$(x^2+x+1{)}^{″}=(2x+1{)}^{\prime }=2;$$

$$(x^2+x+1{)}^{‴}=2^{\prime }=0;$$

...

$$(x^2+x+1{)}^{(15)}=0.$$

From problem 5.201, we write down the derivatives of the sine function:$${\sin }^{(15)}x=\sin (x+15\pi /2)=\sin (x+3\pi /2+6\pi )=\sin (x+3\pi /2)=-\cos x;$$

$${\sin }^{(14)}x=\sin (x+14\pi /2)=\sin (x+\pi +6\pi )=\sin (x+\pi )=-\sin x;$$

$${\sin }^{(13)}x=\sin (x+13\pi /2)=\sin (x+\pi /2+6\pi )=\sin (x+\pi /2)=\cos x.$$

Thus, we find:

$$y^{(15)}=-\cos x(x^2+x+1)+15(-\sin x)(2x+1)+\frac{15\cdot 14}{2}2\cos x=$$ $$=\cos x(-x^2-x-1+209)-15\sin x(2x+1)$$

Answer: $y^{(15)}=\cos x(-x^2-x-1+209)-15\sin x(2x+1)$

10. Find the second-order derivative of the function defined implicitly by $y=1+x{e}^y.$

Solution.

Let's introduce the function $F(x,y):$

$$F(x,y)=y-1-x{e}^y=0.$$

Let's find the first derivative:

$$F_x^{\prime }(x,y)=y^{\prime }-x^{\prime }{e}^y-x(e^y{)}_x^{\prime }=y^{\prime }-e^y-x{e}^y{y}^{\prime }=0.$$ ОFrom here, $y^{\prime }(1-x{e}^y)=e^y\Rightarrow y^{\prime }=\frac{e^y}{1-x{e}^y}.$

Next, we find the second derivative and substitute the found function instead of $y^{\prime }$:

$$F_x^{″}(x,y)=(y^{\prime }-e^y-x{e}^y{y}^{\prime }{)}^{\prime }=y^{″}-e^y{y}^{\prime }-x^{\prime }{e}^y{y}^{\prime }-x(e^y{)}^{\prime }{y}^{\prime }-x{e}^y{y}^{″}=$$

$$=y^{″}-e^y{y}^{\prime }-e^y{y}^{\prime }-x{e}^y{y}^{\prime }{y}^{\prime }-x{e}^y{y}^{″}=$$ $$=y^{″}-2e^y\frac{e^y}{1-x{e}^y}-x{e}^y{\left(\frac{e^y}{1-x{e}^y}\right)}^2-x{e}^y{y}^{″}=0.$$

Now, let's find $y^{″}:$

$$y^{″}-x{e}^y{y}^{″}=2e^y\frac{e^y}{1-x{e}^y}+x{e}^y{\left(\frac{e^y}{1-x{e}^y}\right)}^2=0\Rightarrow$$ $$\Rightarrow y^{″}=\frac{\frac{2e^{2y}}{1-x{e}^y}+\frac{x{e}^{3y}}{(1-x{e}^y{)}^2}}{1-x{e}^y}=\frac{2e^{2y}(1-x{e}^y)+x{e}^{3y}}{(1-x{e}^y{)}^3}=$$ $$=\frac{2e^{2y}-2x{e}^{3y}+x{e}^{3y}}{(1-x{e}^y{)}^3}=\frac{2e^{2y}-x{e}^{3y}}{(1-x{e}^y{)}^3}=e^{2y}\frac{2-x{e}^y}{(1-x{e}^y{)}^3}.$$

Answer: $y^{″}=e^{2y}\frac{2-x{e}^y}{(1-x{e}^y{)}^3}.$

Tags: calculus, derivative, higher-order derivatives, mathematical analysis