heureka

Find dy/dx

Y^x+X^X+X^y= a^b

\(\begin{array}{|rcll|} \hline y^x+x^x+x^y &=& a^b \\ y^x+x^x+x^y -a^b &=& 0 \\ \hline \end{array} \)

Formula:

\(\begin{array}{|rcll|} \hline \frac{dy}{dx} = - \dfrac{F_x}{F_y} \\ \hline \end{array} \)

\(\mathbf{F_x =\ ?} \begin{array}{|rcll|} \hline F_x &=& \frac{d }{dx}y^x + \frac{d }{dx}x^x + \frac{d }{dx}x^y + \frac{d }{dx}a^b \\\\ h &=& y^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(y) \\ \frac{h'}{h} &=& \ln(y) \\ h' &=& h\cdot \ln(y) \\ h' &=& y^x\cdot \ln(y) \\ \mathbf{ \frac{d }{dx}y^x } & \mathbf{=} & \mathbf{ y^x\cdot \ln(y) } \\\\ h &=& x^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(x) \\ \frac{h'}{h} &=& x\cdot \frac{1}{x} + 1\cdot \ln(x) \\ h' &=& h \cdot \Big( 1+\ln(x) \Big) \\ \mathbf{ \frac{d }{dx}x^x } & \mathbf{=} & \mathbf{ x^x \cdot \Big( 1+ \ln(x) \Big) } \\\\ h &=& x^y \quad & | \quad \ln() \\ \ln(h) &=& y\cdot \ln(x) \\ \frac{h'}{h} &=& \frac{y}{x} \\ h' &=& h \cdot \frac{y}{x} \\ h' &=& x^y \cdot \frac{y}{x} \\ \mathbf{ \frac{d }{dx}x^y } & \mathbf{=} & \mathbf{ x^y \cdot \frac{y}{x} } \\\\ \mathbf{ \frac{d }{dx}a^b } & \mathbf{=} & \mathbf{ 0 } \\ \hline \end{array} \)

\(\mathbf{F_y =\ ?} \begin{array}{|rcll|} \hline F_y &=& \frac{d }{dy}y^x + \frac{d }{dy}x^x + \frac{d }{dy}x^y +\frac{d }{dy}a^b \\\\ h &=& y^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(y) \\ \frac{h'}{h} &=& \frac{x}{y} \\ h' &=& h\cdot \frac{x}{y} \\ h' &=& y^x\cdot \frac{x}{y} \\ \mathbf{ \frac{d }{dy}y^x } & \mathbf{=} & \mathbf{ y^x\cdot \frac{x}{y} } \\\\ \mathbf{ \frac{d }{dy}x^x } & \mathbf{=} & \mathbf{ 0 } \\ h &=& x^y \quad & | \quad \ln() \\ \ln(h) &=& y\cdot \ln(x) \\ \frac{h'}{h} &=& \ln(x) \\ h' &=& h \cdot \ln(x) \\ h' &=& x^y \cdot \ln(x) \\ \mathbf{ \frac{d }{dy}x^y } & \mathbf{=} & \mathbf{ x^y \cdot \ln(x) } \\\\ \mathbf{ \frac{d }{dy}a^b } & \mathbf{=} & \mathbf{ 0 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \frac{dy}{dx} &=& - \dfrac{F_x}{F_y} \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^y \cdot \frac{y}{x} + 0 } { y^x\cdot \frac{x}{y} + 0 + x^y \cdot \ln(x) + 0 } \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^y \cdot \frac{y}{x} } { y^x\cdot \frac{x}{y} + x^y \cdot \ln(x) } \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^{y-1} \cdot y } { y^{x-1}\cdot x + x^y \cdot \ln(x) } \\\\ \mathbf{ \frac{dy}{dx} } & \mathbf{=} & \mathbf{ - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^{y-1} \cdot y } { x^y \cdot \ln(x) + y^{x-1}\cdot x } } \\ \hline \end{array}\)

f(x) = (3x-6)/x

\(\begin{array}{|rcll|} \hline f(x) &=& (3x-6)/x \\ f(x) &=& (3x-6)\cdot x^{-1} \\ f'(x) &=& (3x-6)\cdot (-1)\cdot x^{-2} + 3 \cdot x^{-1} \\ f'(x) &=& \frac{6-3x}{x^2} + \frac{3}{x} \cdot \frac{x}{x} \\ f'(x) &=& \frac{6-3x+3x}{x^2} \\ f'(x) &=& \frac{6}{x^2} \\ \hline \end{array}\)

Simple rules of differentiation

Why is this true?

Differentiation of xⁿ

f(x) = xⁿ

f'(x) = nx^(n-1)

difference quotient:

\(\begin{array}{|rcll|} \hline \frac{\Delta y}{\Delta x} &=& \frac{(x+h)^n-x^n}{h} \\ \frac{\Delta y}{\Delta x} &=& \frac{ \binom n0 x^n + \binom n1 x^{n-1}h+ \binom n2 x^{n-2}h^2\ldots + \binom nn h^n -x^n }{h} \quad & | \quad \binom n0 = 1 \\ \frac{\Delta y}{\Delta x} &=& \frac{ x^n + \binom n1 x^{n-1}h+ \binom n2 x^{n-2}h^2\ldots + \binom nn h^n -x^n }{h} \\ \frac{\Delta y}{\Delta x} &=& \frac{ \binom n1 x^{n-1}h+ \binom n2 x^{n-2}h^2\ldots + \binom nn h^n }{h} \\ \frac{\Delta y}{\Delta x} &=& \binom n1 x^{n-1}+ \binom n2 x^{n-2}h^1\ldots + \binom nn h^{n-1} \quad & | \quad \binom n1 = n \\ \frac{\Delta y}{\Delta x} &=& n x^{n-1}+ \binom n2 x^{n-2}h\ldots + \binom nn h^{n-1} \\ \hline \end{array} \)

differential quotient:

\(\begin{array}{|rcll|} \hline \frac{\delta y}{\delta x} =f'(x) &=& \lim \limits_{h\to 0} \left( n x^{n-1}+ \binom n2 x^{n-2}h\ldots + \binom nn h^{n-1} \right) \\ f'(x) &=& n x^{n-1}+ \binom n2 x^{n-2}\times 0 \ldots + \binom nn \times 0^{n-1} \\ f'(x) &=& n x^{n-1} + 0 \ldots +0 \\ f'(x) &=& n x^{n-1} \\ \hline \end{array} \)

A cup has a base radius of 3cm,
a height of 15cm and a top radius of 5cm,
calculate the volume
i got 769.69ml, just wanna know if im right or not

Volume cup:  \(V = \frac13 \pi \cdot 5^2 \cdot H - \frac13 \pi \cdot 3^2 \cdot ( H - 15 ) \)

Height H of the cone: \(\frac{H-15}{3} = \frac{H}{5}\)

\(\begin{array}{|rcll|} \hline \frac{H-15}{3} &=& \frac{H}{5} \\ 5\cdot(H-15) &=& 3H \\ 5H-75 &=& 3H \\ 2H &=& 75 \\ H &=& \frac{75}{2} \\ \mathbf{H} &\mathbf{=}& \mathbf{37.5} \\\\ H-15 &=& 37.5 - 15 \\ \mathbf{H-15} &\mathbf{=}& \mathbf{22.5} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline V &=& \frac13 \pi \cdot 5^2 \cdot H - \frac13 \pi \cdot 3^2 \cdot ( H - 15 ) \\\\ &=& \frac13 \pi \cdot 5^2 \cdot 37.5 - \frac13 \pi \cdot 3^2 \cdot 22.5 \\\\ &=& \frac13 \pi \cdot ( 5^2 \cdot 37.5 - 3^2 \cdot 22.5 ) \\\\ &=& \frac13 \pi \cdot ( 25 \cdot 37.5 - 9 \cdot 22.5 ) \\\\ &=& \frac13 \pi \cdot ( 937.5 - 202.5 ) \\\\ &=& \frac13 \pi \cdot 735 \\\\ &=& 245 \pi \\ &=& 769.690200129 \ cm^3 \\ \mathbf{V} &\mathbf{=}& \mathbf{769.69\ldots \ ml} \\ \hline \end{array}\)

The function f(x,y) accepts an ordered pair as input and gives another ordered pair as output.
It is defined according to the following rules:
If x > 4, f(x,y) = (x - 4,y).
If x less equal 4 but y > 4, f(x,y) = (x,y - 4).
Otherwise, f(x,y) = (x + 5, y + 6).
A robot starts by moving to the point (1,1).
Every time it arrives at a point (x,y),
it applies f to that point and then moves to f(x,y).
If the robot runs forever,
how many different points will it visit?

y=(x+2)(x-1)² differentiate in respect to x

\(y=(x+2)(x-1)^2 \\ y=(x+2)(x^2-2x+1) \\ y=x^3-2x^2+x+2x^2-4x+3 \\ y = x^3-3x+3\\ \mathbf{y' = 3x^2-3 }\)

lim n→∞{(1− 3/n)^3n +3*n'te√3n}

\(\lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) }= \ ? \)

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) } + \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } \\ \hline \end{array} \)

\(y=\lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) }= \ ?\)

\(\begin{array}{|rcll|} \hline \ln(y) &=& \ln\Big( \lim \limits_{x\to 0} \Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) \Big) \\ &=& \lim \limits_{x\to 0} \Big( \ln \Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) \Big) \\ &=& \lim \limits_{x\to 0} \Big( 3n\ln \left( 1- \frac{3}{n} \right) \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{3\ln \left( 1- \frac{3}{n} \right)} {n^{-1}} \Big) \\\\ && \text{Regel von de l’Hospital anwenden} \\ && \text{Die 1. Ableitung von } 3\cdot \ln(1-3\cdot n^{-1}) \text{ lautet } 3 \cdot \frac{ [(-3)(-1)\cdot n^{-2} ] } {1-\frac{3}{n} } \\\\ &=& \lim \limits_{x\to 0} \Big( \frac{3 \cdot \frac{ [(-3)(-1)\cdot n^{-2} ] } {1-\frac{3}{n} } } {(-1)\cdot n^{-2}} \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ 3 \cdot[(-3)(-1)\cdot n^{-2} ] } { (1-\frac{3}{n})(-1)\cdot n^{-2} } \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ 3 \cdot(-3)} { (1-\frac{3}{n}) } \Big) \\ &=& \lim \limits_{x\to 0} \Big( \frac{ -9 } { (1-\frac{3}{n}) } \Big) \\ \ln(y) &=& \frac{ -9 } { 1-0 } \\ \ln(y) &=& -9 \\ e^{\ln(y)} = y &=& e^{-9}\\ y &=& \lim \limits_{x\to 0} {\Big( \left( 1- \frac{3}{n} \right)^{3n} \Big) }= e^{-9} \\ \hline \end{array}\)

\(\lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } = \ ?\)

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot (3n)^{\frac{1}{n}} \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot e^{\ln \left((3n)^{\frac{1}{n}} \right) } \Big) } \\ &=& \lim \limits_{x\to 0} {\Big( 3 \cdot e^{ \frac{\ln(3n)} {n} } \Big) } \\ &=& 3 \cdot e^{ \lim \limits_{x\to 0} {\Big( \frac{\ln(3n)} {n} } \Big) } \\ && \text{Regel von de l’Hospital anwenden} \\ && \text{Die 1. Ableitung von } \frac{\ln(3n)} {n} \text{ lautet } \frac{ \frac{3}{3n} } {1} = \frac{3}{3n} = \frac{1}{n} \\\\ &=& 3 \cdot e^{ \lim \limits_{x\to 0} {\Big(\frac{1}{n}} \Big) } \\ &=& 3 \cdot e^0 \\ &=& 3 \cdot 1 \\ &=& 3 \\ \lim \limits_{x\to 0} {\Big( 3 \cdot \sqrt[n]{3n} \Big) } &=& 3 \\ \hline \end{array} \)

Somit erhalten wir insgesamt:

\(\lim \limits_{x\to 0} { \Big( \left( 1- \frac{3}{n} \right)^{3n} +3 \cdot \sqrt[n]{3n} \Big) } = e^{-9} + 3\)

simplify this expression?

\(\dfrac{2^{-1}a^2b^{-3}}{4a^3b^{-5}}\)

\(\begin{array}{|rcll|} \hline && \dfrac{2^{-1}a^2b^{-3}}{4a^3b^{-5}} \\\\ &=& \dfrac{2^{-1}a^2b^{-3}}{2^2a^3b^{-5}} \\\\ &=& \dfrac{2^{-1}}{2^2}\cdot \dfrac{a^2}{a^3} \cdot \dfrac{b^{-3}}{b^{-5}} \\\\ &=& \dfrac{1}{2^22^{1}}\cdot \dfrac{1}{a^3a^{-2}} \cdot \dfrac{b^{-3}b^{5}}{1} \\\\ &=& \dfrac{1}{2^{2+1}}\cdot \dfrac{1}{a^{3-2}} \cdot \dfrac{b^{-3+5}}{1} \\\\ &=& \dfrac{1}{2^{3}}\cdot \dfrac{1}{a^{1}} \cdot \dfrac{b^{2}}{1} \\\\ &=& \dfrac{1}{2^{3}}\cdot \dfrac{1}{a} \cdot \dfrac{b^{2}}{1} \\\\ &=& \dfrac{b^{2}}{2^{3}a} \\\\ &=& \dfrac{b^{2}}{8a} \\ \hline \end{array}\)

Compute

\(1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\)

Let \(r =\frac12\)

\(\begin{array}{|rcll|} \hline s_n &=& 1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} \\\\ && r = \frac12 \\\\ \hline s_n &=& 1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \dots + n \cdot r^n \\ rs_n &=& \qquad \quad 1\cdot r^2 + 2\cdot r^3 + \dots + (n-1) \cdot r^n + n\cdot r^{n+1} \\ \hline s_n -r\cdot s_n &=& r+r^2+r^3+ \dots +r^n-n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ \hline && S_n = r+r^2+r^3+ \dots +r^n \\ && rS_n = \quad r^2+r^3 + \dots + r^{n+1} \\ \hline && S_n -r\cdot S_n = r - r^{n+1} \\ && S_n\cdot (1-r) = r - r^{n+1} \\ && S_n = \frac{r - r^{n+1}}{1-r} \\ \hline s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \\ s_n &=& \frac{1}{1-r} \cdot \Big( \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \Big) \\\\ && \frac{1}{1-r} = 2 \\\\ s_n &=& 2 \cdot \Big( 2\cdot(r - r^{n+1}) -n\cdot r^{n+1} \Big) \\ s_n &=& 2 \cdot ( 2\cdot r - 2\cdot r^{n+1} - n\cdot r^{n+1} ) \\ s_n &=& 2 \cdot r\cdot ( 2 - 2\cdot r^n - n\cdot r^n ) \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 \cdot r\cdot \Big( 2 - (2+n)\cdot r^n \Big)} \quad & | \quad r=\frac12 \\ \\ s_n & = & 2 \cdot \frac12 \cdot \Big( 2 - (2+n)\cdot (\frac12)^n \Big) \\ s_n & = & 2 - (2+n)\cdot \frac{1}{2^n} \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 - \frac{2+n}{2^n} } \\ \hline \end{array}\)

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84.

What is the 23rd term?

Formula arithmetic sequence \(a_n = a_1 +(n-1)d\)

\(\begin{array}{|lrcll|} \hline & a_{23} &=& a_1 + (23-1)d \\ (1) & a_{23} &=& a_1 + 22d \\\\ & a_5 = 9 &=& a_1 + (5-1)d \\ (2) & 9 &=& a_1 + 4d \\\\ & a_{32}=-84 &=& a_1 + (32-1)d \\ (3) & -84 &=& a_1 + 31d \\ \hline (1)-(2): & a_{23} - 9 &=& 18d \\ (1)-(3): & a_{23} + 84 &=& -9d \\ \hline & \frac{a_{23} - 9}{a_{23} + 84} &=& \frac{18d}{-9d } \\ & \frac{a_{23} - 9}{a_{23} + 84} &=& -2 \\ & a_{23} - 9 &=& -2(a_{23} + 84) \\ & a_{23} - 9 &=& -2a_{23} -168 \\ & 3a_{23} &=& 9 -168 \\ & 3a_{23} &=& -159 \\ & a_{23} &=& -\frac{159}{3} \\\\ & \mathbf{ a_{23} } & \mathbf{=} & \mathbf{-53} \\ \hline \end{array}\)