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
$$f'(x_0)=\lim\limits_{\Delta x\rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}.$$
Key formulas for calculating derivatives.
$c'=0,\quad c=const;$
$(x^\alpha)'=\alpha x^{\alpha-1},$ $x\in\mathbb{R}, \alpha\neq 0;$
$(a^x)'=a^x\ln a,\quad a>0, a\neq 1, x\in \mathbb{R};$
$(e^x)'=e^x;$
$(\log_a x)'=\frac{1}{x\ln a}, \quad x>0;$
$(\log_a|x|)'=\frac{1}{x\ln a},\quad x\neq 0;$
$(\ln x)'=\frac{1}{x},\quad x>0;$
$(\sin x)'=\cos x, \quad x\in \mathbb{R};$
$(\cos x)'=-\sin x\quad x\in \mathbb{R};$
$(tg x)'=\frac{1}{\cos^2 x},$ $x\neq\frac{\pi}{2}(2n+1),\,\, n\in\mathbb{Z};$
$(ctg x)'=-\frac{1}{\sin^2 x},$ $x\neq\pi n,\,\, n\in\mathbb{Z};$
$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}},\quad |x|<1;$
$(\arccos x)'=-\frac{1}{\sqrt{1-x^2}},\quad |x|<1;$
$(arctg x)'=\frac{1}{1+x^2},\quad x\in\mathbb{R};$
$(arcctg x)'=-\frac{1}{1+x^2},\quad x\in\mathbb{R};$
Rules for computing derivatives.
1) Let $f=c_1 f_1+c_2f_2+...+c_nf_n.$ Then $f'=c_1f'_1+c_2f'_2+...+c_nf'_n.$
2) Let $f=f_1\cdot f_2.$ Then $f'=f'_1f_2+f_1f'_2.$
3) If $f=f_1/f_2,$ then $f'=\frac{f'_1f_2-f'_2f_1}{f_2^2}.$
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 $z=\varphi(x)=g(f(x)),$ also has a derivative at $x_0,$ and $\varphi'(x_0)=g'(y_0)f'(x_0).$
Examples.
Find the derivative of the function $y(x)$.
1) $y = x^3 + x^2 + x + 1$
Solution.
$y'=3x^2+2x+1.$
2) ${ y=7x^{13}+13x^{-7}}$
Solution.
$y'=7\cdot 13x^{12}+13({-7})x^{-8}=91x^{12}-91x^{-8}=91(x^{12}-x^{-8}).$
3) ${y=\frac{x^2-5x+6}{x^2+x+7}}$
Solution.
$$y'=\frac{(x^2-5x+6)'(x^2+x+7)-(x^2+x+7)'(x^2-5x+6)}{(x^2+x+7)^2}=$$ $$=\frac{(2x-5)(x^2+x+7)-(2x+1)(x^2-5x+6)}{(x^2+x+7)^2}=$$ $$\frac{2x^3+2x^2+14x-5x^2-5x-35-2x^3+10x^2-12x-x^2+5x-6}{(x^2+x+7)^2}=$$ $$=\frac{6x^2+2x-41}{(x^2+x+7)^2}.$$
4) ${y=(x+1)tg x;}$
Solution.
$y'=(x+1)'tg x+(x+1)(tg x)'=tg x+\frac{x+1}{\cos^2 x};$
5) ${ y=(a+bx)^\alpha}$
Solution.
$y'=\alpha(a+bx)^{\alpha-1}(a+bx)'=$ $\alpha(a+bx)^{\alpha-1}b;$
6) ${y=\sqrt[3]{\frac{1-x^3}{1+x^3}}.}$
Solution.
$$y'=\frac{1}{3}\left(\frac{1-x^3}{1+x^3}\right)^{-\frac{2}{3}}\frac{(-3x^2)(1+x^3)-3x^2(1-x^3)}{(1+x^3)^2}$$
7) ${y=\ln\ln(x/2)}$
Solution.
$$y'=\frac{1}{\ln(x/2)}\cdot(\ln(x/2))'=\frac{1}{\ln(x/2)}\cdot\frac{2}{x}\cdot\frac{1}{2}=\frac{1}{x\log(x/2)}.$$
8) $y=\ln\frac{x^2+a}{\sqrt{x^4+b^2}}+\frac{a}{b}arctg\frac{x^2}{b}$
Solution.
$$y'=\frac{\sqrt{x^4+b^2}}{x^2+a}\cdot\left(\frac{x^2+a}{\sqrt{x^4+b^2}}\right)'+\frac{a}{b}\frac{1}{1+\frac{x^4}{b^2}}\cdot\left(\frac{x^2}{b}\right)'=$$
$$\frac{\sqrt{x^4+b^2}}{x^2+a}\cdot\frac{2x\sqrt{x^4+b^2}-1/2(x^4+b^2)^{-1/2}\cdot4x^3(x^2+a)}{x^4+b^2}+\frac{a}{b}\frac{1}{1+\frac{x^4}{b^2}}\cdot\frac{2x}{b}=$$
$$=\frac{2x(x^4+b^2)-2x^3(x^2+a)+2xa(x^2+a)}{(x^2+a)(x^4+b^2)}=\frac{2x(a^2+b^2)}{(x^2+a)(x^4+b^2)}.$$
9) ${y=x^x}$
Solution.
$y'=(x^x)'=(e^{\ln x^x})'=(e^{x\ln x})'=e^{x\ln x}(x\ln x)'=e^{x\ln x}(\ln x+x\cdot\frac{1}{x})=$ $=x^x(\ln x+1).$
10) ${y=\log_x 7}$
Solution.
$$y'=(\log_x 7)'=\left(\frac{1}{\log_7 x}\right)'=(\log_7^{-1} x)'=-\log_7^{-2}x\cdot\frac{1}{x\ln 7}=$$
$$-\frac{1}{(\frac{\ln x}{\ln 7})^2}\cdot\frac{1}{x\ln 7}=-\frac{1}{x\ln x\cdot\frac{\ln x}{\ln 7}}=-\frac{1}{x\ln x\log_7 x}.$$
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 $\frac{dx(t_0)}{dt} \neq 0$, then the function $y=f(x)$ at the point $x_0$ also has a derivative, which can be given by the formula:
$\frac{df(x_0)}{dx}=\frac{\frac{dy(t_0)}{dt}}{\frac{dx(t_0)}{dt}},$ $y'_x(x_0)=\frac{y'_t(t_0)}{x'_t(t_0)}.$
Examples.
1) ${ x=\sin^2 t,\,\, y=\cos^2 t,\quad t\in(0,\pi/2). }$
Solution.
$x'_t=2\sin t\cdot(\sin t)'=2\sin t\cdot\cos t;$
$y'_t=2\cos t\cdot(\cos t)'=2\cos t\cdot(-\sin t).$
$$y'_x=\frac{2\cos t\cdot(-\sin t)}{2\sin t\cdot\cos t}=-1.$$
2) ${ x=e^{-t}; \,\, y=t^3 \quad t\in (-\infty;\, +\infty) .}$
Solution.
$$y'_x=\frac{3t^2}{-e^{-t}}=-3t^2e^t.$$
3) ${ x=a\cos t;\,\, y=b\sin t,\quad t\in(0,\pi). }$
Solution.
$$y'_x=\frac{b\cos t}{-a\sin t}-\frac{b}{a}ctg t.$$
4) ${ x=(t^3-2t^2+3t-4)e^t,\,\, y=(t^3-2t^2+4t-4)e^t. }$
Solution.
$$y'_x=\frac{(3t^2-4t+4)e^t+e^t(t^3-2t^2+4t-4)}{(3t^2-4t+3)e^t+e^t(t^3-2t^2+3t-4)}=$$ $$\frac{t^3+t^2}{t^3+t^2-t-1}=\frac{t^2(t+1)}{t^2(t+1)-(t+1)}=\frac{t^2}{t^2-1}.$$
5) $x=ctg 2t,\,\, y=\frac{2\cos 2t-1}{2\cos t}\quad t\in (0, \pi/2).$ $x'_y -?$
Solution.
$$x'_y=-\frac{2}{\sin^2 2t}\cdot\frac{4\cos^2 t}{-4\sin 2t\cdot2\cos t+2\sin t(2\cos 2t-1))}=$$ $$\frac{-1}{\sin^2 t(-8\sin t\cos^2 t+\sin t(2(1-2\sin^2 t)-1))}=$$
$$\frac{-1}{-8\sin^3 t\cos^2 t-4\sin^5 t+2\sin^3 t-\sin^3 t}=$$
$$\frac{1}{\sin^3 t(8\cos^2 t+4\sin^2t-2+1)}=\frac{1}{\sin^3 t(4\cos^2 t+3)}.$$
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 $y'(x)$ can be found from the equation $\frac{d}{dx}F(x,y)=0$.
Examples.
Find the derivative of the function $y'(x)$.
1) ${ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 }$
Solution.
$$\frac{d}{dx}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)=0$$
$$\frac{2x}{a^2}+\frac{2y}{b^2}y'=0, \Rightarrow\quad y'=-\frac{b^2 x}{a^2 y}.$$
2) ${ y^5+y^3+y-x=0 }$
Solution.
$$\frac{d}{dx}(y^5+y^3+y-x)=0 \Rightarrow 5y^4y'+3y^2y'+y'-1=0\,\,\Rightarrow\,\,$$ $$\Rightarrow y'=\frac{1}{5y^4+3y^2+1}.$$
3) $ y-x=\varepsilon\sin y $
Solution.
$$\frac{d}{dx}(y-x-\varepsilon\sin y)=0\Rightarrow\,\, y'-1-\varepsilon\cos y\cdot y'=0 \Rightarrow y'=\,\frac{1}{1-\varepsilon\cos y}.$$
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