The derivative of a function. Calculating first-order derivatives.

Definition: The first-order derivative of a function $y=f(x)$ at the point $x_0$ is defined as the limit

$$f^{\prime }(x_0)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x_0+\mathrm{\Delta }x)-f(x_0)}{\mathrm{\Delta }x}.$$

Key formulas for calculating derivatives.

$c^{\prime }=0,c=const;$

$(x^{\alpha }{)}^{\prime }=\alpha x^{\alpha -1},$ $x\in \mathbb{R},\alpha \ne 0;$

$(a^x{)}^{\prime }=a^x\ln a,a>0,a\ne 1,x\in \mathbb{R};$

$(e^x{)}^{\prime }=e^x;$

$({\log }_{a}x{)}^{\prime }=\frac{1}{x\ln a},x>0;$

$({\log }_a|x|{)}^{\prime }=\frac{1}{x\ln a},x\ne 0;$

$(\ln x{)}^{\prime }=\frac{1}{x},x>0;$

$(\sin x{)}^{\prime }=\cos x,x\in \mathbb{R};$

$(\cos x{)}^{\prime }=-\sin xx\in \mathbb{R};$

$(tgx{)}^{\prime }=\frac{1}{{\cos }^{2}x},$ $x\ne \frac{\pi }{2}(2n+1),n\in \mathbb{Z};$

$(ctgx{)}^{\prime }=-\frac{1}{{\sin }^{2}x},$ $x\ne \pi n,n\in \mathbb{Z};$

$(\arcsin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}},|x|<1;$

$(\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}},|x|<1;$

$(arctgx{)}^{\prime }=\frac{1}{1+x^2},x\in \mathbb{R};$

$(arcctgx{)}^{\prime }=-\frac{1}{1+x^2},x\in \mathbb{R};$

Rules for computing derivatives.

1) Let $f=c_1{f}_1+c_2{f}_2+...+c_n{f}_n.$ Then $f^{\prime }=c_1{f}_1^{\prime }+c_2{f}_2^{\prime }+...+c_n{f}_n^{\prime }.$

2) Let $f=f_1\cdot f_2.$ Then $f^{\prime }=f_1^{\prime }{f}_2+f_1{f}_2^{\prime }.$

3) If $f=f_1/f_2,$ then $f^{\prime }=\frac{f_1^{\prime }{f}_2-f_2^{\prime }{f}_1}{f_2^2}.$

4) If the function $y=f(x)$ has a derivative at the point $x_0,$ and the function $z=g(y)$ has a derivative at the point $y_0=f(x_0),$ then the composite function $z=\phi (x)=g(f(x)),$ also has a derivative at $x_0,$ and ${\phi }^{\prime }(x_0)=g^{\prime }(y_0)f^{\prime }(x_0).$

Examples.

Find the derivative of the function $y(x)$.

1) $y=x^3+x^2+x+1$

Solution.

$y^{\prime }=3x^2+2x+1.$

2) $y=7x^{13}+13x^{-7}$

Solution.

$y^{\prime }=7\cdot 13x^{12}+13(-7)x^{-8}=91x^{12}-91x^{-8}=91(x^{12}-x^{-8}).$

3) $y=\frac{x^2-5x+6}{x^2+x+7}$

Solution.

$$y^{\prime }=\frac{(x^2-5x+6{)}^{\prime }(x^2+x+7)-(x^2+x+7{)}^{\prime }(x^2-5x+6)}{(x^2+x+7{)}^2}=$$ $$=\frac{(2x-5)(x^2+x+7)-(2x+1)(x^2-5x+6)}{(x^2+x+7{)}^2}=$$ $$\frac{2x^3+2x^2+14x-5x^2-5x-35-2x^3+10x^2-12x-x^2+5x-6}{(x^2+x+7{)}^2}=$$ $$=\frac{6x^2+2x-41}{(x^2+x+7{)}^2}.$$

4) $y=(x+1)tgx;$

Solution.

$y^{\prime }=(x+1{)}^{\prime }tgx+(x+1)(tgx{)}^{\prime }=tgx+\frac{x+1}{{\cos }^{2}x};$

5) $y=(a+bx{)}^{\alpha }$

Solution.

$y^{\prime }=\alpha (a+bx{)}^{\alpha -1}(a+bx{)}^{\prime }=$ $\alpha (a+bx{)}^{\alpha -1}b;$

6) $y=\sqrt[3]{\frac{1-x^3}{1+x^3}}.$

Solution.

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

7) $y=\ln \ln (x/2)$

Solution.

$$y^{\prime }=\frac{1}{\ln (x/2)}\cdot (\ln (x/2){)}^{\prime }=\frac{1}{\ln (x/2)}\cdot \frac{2}{x}\cdot \frac{1}{2}=\frac{1}{x\log (x/2)}.$$

8) $y=\ln \frac{x^2+a}{\sqrt{x^4+b^2}}+\frac{a}{b}arctg\frac{x^2}{b}$

Solution.

$$y^{\prime }=\frac{\sqrt{x^4+b^2}}{x^2+a}\cdot {\left(\frac{x^2+a}{\sqrt{x^4+b^2}}\right)}^{\prime }+\frac{a}{b}\frac{1}{1+\frac{x^4}{b^2}}\cdot {\left(\frac{x^2}{b}\right)}^{\prime }=$$

$$\frac{\sqrt{x^4+b^2}}{x^2+a}\cdot \frac{2x\sqrt{x^4+b^2}-1/2(x^4+b^2{)}^{-1/2}\cdot 4x^3(x^2+a)}{x^4+b^2}+\frac{a}{b}\frac{1}{1+\frac{x^4}{b^2}}\cdot \frac{2x}{b}=$$

$$=\frac{2x(x^4+b^2)-2x^3(x^2+a)+2xa(x^2+a)}{(x^2+a)(x^4+b^2)}=\frac{2x(a^2+b^2)}{(x^2+a)(x^4+b^2)}.$$

9) $y=x^x$

Solution.

$y^{\prime }=(x^x{)}^{\prime }=(e^{\ln x^x}{)}^{\prime }=(e^{x\ln x}{)}^{\prime }=e^{x\ln x}(x\ln x{)}^{\prime }=e^{x\ln x}(\ln x+x\cdot \frac{1}{x})=$ $=x^x(\ln x+1).$

10) $y={\log }_{x}7$

Solution.

$$y^{\prime }=({\log }_{x}7{)}^{\prime }={\left(\frac{1}{{\log }_{7}x}\right)}^{\prime }=({\log }_7^{-1}x{)}^{\prime }=-{\log }_7^{-2}x\cdot \frac{1}{x\ln 7}=$$

$$-\frac{1}{(\frac{\ln x}{\ln 7}{)}^2}\cdot \frac{1}{x\ln 7}=-\frac{1}{x\ln x\cdot \frac{\ln x}{\ln 7}}=-\frac{1}{x\ln x{\log }_{7}x}.$$

The derivative of a function defined parametrically.

Let functions $x=x(t)$ and $y=y(t)$ be defined in some neighborhood of $t_0$, and parametrically define the function $y=f(x)$ in the neighborhood of $x_0=x(t_0)$. Then, if $x(t)$ and $y(t)$ have derivatives at the point $t_0$ and if $\frac{dx(t_0)}{dt}\ne 0$, then the function $y=f(x)$ at the point $x_0$ also has a derivative, which can be given by the formula:

$\frac{df(x_0)}{dx}=\frac{\frac{dy(t_0)}{dt}}{\frac{dx(t_0)}{dt}},$ $y_x^{\prime }(x_0)=\frac{y_t^{\prime }(t_0)}{x_t^{\prime }(t_0)}.$

Examples.

1) $x={\sin }^{2}t,y={\cos }^{2}t,t\in (0,\pi /2).$

Solution.

$x_t^{\prime }=2\sin t\cdot (\sin t{)}^{\prime }=2\sin t\cdot \cos t;$

$y_t^{\prime }=2\cos t\cdot (\cos t{)}^{\prime }=2\cos t\cdot (-\sin t).$

$$y_x^{\prime }=\frac{2\cos t\cdot (-\sin t)}{2\sin t\cdot \cos t}=-1.$$

2) $x=e^{-t};y=t^{3}t\in (-\mathrm{\infty };+\mathrm{\infty }).$

Solution.

$$y_x^{\prime }=\frac{3t^2}{-e^{-t}}=-3t^2{e}^t.$$

3) $x=a\cos t;y=b\sin t,t\in (0,\pi ).$

Solution.

$$y_x^{\prime }=\frac{b\cos t}{-a\sin t}-\frac{b}{a}ctgt.$$

4) $x=(t^3-2t^2+3t-4)e^t,y=(t^3-2t^2+4t-4)e^t.$

Solution.

$$y_x^{\prime }=\frac{(3t^2-4t+4)e^t+e^t(t^3-2t^2+4t-4)}{(3t^2-4t+3)e^t+e^t(t^3-2t^2+3t-4)}=$$ $$\frac{t^3+t^2}{t^3+t^2-t-1}=\frac{t^2(t+1)}{t^2(t+1)-(t+1)}=\frac{t^2}{t^2-1}.$$

5) $x=ctg2t,y=\frac{2\cos 2t-1}{2\cos t}t\in (0,\pi /2).$ $x_y^{\prime }-?$

Solution.

$$x_y^{\prime }=-\frac{2}{{\sin }^{2}2t}\cdot \frac{4{\cos }^{2}t}{-4\sin 2t\cdot 2\cos t+2\sin t(2\cos 2t-1))}=$$ $$\frac{-1}{{\sin }^{2}t(-8\sin t{\cos }^{2}t+\sin t(2(1-2{\sin }^{2}t)-1))}=$$

$$\frac{-1}{-8{\sin }^{3}t{\cos }^{2}t-4{\sin }^{5}t+2{\sin }^{3}t-{\sin }^{3}t}=$$

$$\frac{1}{{\sin }^{3}t(8{\cos }^{2}t+4{\sin }^{2}t-2+1)}=\frac{1}{{\sin }^{3}t(4{\cos }^{2}t+3)}.$$

The derivative of a function defined implicitly.

If a differentiable function $y=y(x)$ defined on some interval is implicitly given by the equation $F(x,y)=0$, then its derivative $y^{\prime }(x)$ can be found from the equation $\frac{d}{dx}F(x,y)=0$.

Examples.

Find the derivative of the function $y^{\prime }(x)$.

1) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Solution.

$$\frac{d}{dx}(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1)=0$$

$$\frac{2x}{a^2}+\frac{2y}{b^2}{y}^{\prime }=0,\Rightarrow y^{\prime }=-\frac{b^{2}x}{a^{2}y}.$$

2) $y^5+y^3+y-x=0$

Solution.

$$\frac{d}{dx}(y^5+y^3+y-x)=0\Rightarrow 5y^4{y}^{\prime }+3y^2{y}^{\prime }+y^{\prime }-1=0\Rightarrow$$ $$\Rightarrow y^{\prime }=\frac{1}{5y^4+3y^2+1}.$$

3) $y-x=\epsilon \sin y$

Solution.

$$\frac{d}{dx}(y-x-\epsilon \sin y)=0\Rightarrow y^{\prime }-1-\epsilon \cos y\cdot y^{\prime }=0\Rightarrow y^{\prime }=\frac{1}{1-\epsilon \cos y}.$$

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