Question

o). express $\frac{7x}{x^{2}-7x + 10}$ as a sum of its partial fractions.
p. determine the solutions for the equation $z^{3}=-8$. give your answers in polar form $r cis\theta$, where angles are in radians.
section three
derivatives
a. use power rule to differentiate $y = x^{4}-5$
b. differentiate: i). $f(x)=cos5x$

Explanation:

Step1: Set up partial - fraction decomposition

Let $\frac{7x}{x^{2}-7x + 10}=\frac{7x}{(x - 5)(x - 2)}=\frac{A}{x - 5}+\frac{B}{x - 2}$. Then $7x=A(x - 2)+B(x - 5)$.

Step2: Find the value of A

Set $x = 5$, we get $7\times5=A(5 - 2)+B(5 - 5)$, so $35 = 3A$, and $A=\frac{35}{3}$.

Step3: Find the value of B

Set $x = 2$, we get $7\times2=A(2 - 2)+B(2 - 5)$, so $14=-3B$, and $B=-\frac{14}{3}$.
So $\frac{7x}{x^{2}-7x + 10}=\frac{35/3}{x - 5}-\frac{14/3}{x - 2}$.

Step4: Solve the complex - number equation $z^{3}=-8$

First, write $-8$ in polar form. We have $-8 = 8(\cos\pi+i\sin\pi)$.
By De - Moivre's theorem, if $z^{n}=r(\cos\theta+i\sin\theta)$, then $z = r^{\frac{1}{n}}(\cos\frac{\theta + 2k\pi}{n}+i\sin\frac{\theta + 2k\pi}{n})$, where $k = 0,1,\cdots,n - 1$.
For $z^{3}=8(\cos\pi+i\sin\pi)$, $r = 8$, $n = 3$, $\theta=\pi$.
When $k = 0$:
$z_1=8^{\frac{1}{3}}(\cos\frac{\pi+0}{3}+i\sin\frac{\pi + 0}{3})=2(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=2\mathrm{cis}\frac{\pi}{3}$
When $k = 1$:
$z_2=8^{\frac{1}{3}}(\cos\frac{\pi + 2\pi}{3}+i\sin\frac{\pi+2\pi}{3})=2(\cos\pi+i\sin\pi)=2\mathrm{cis}\pi$
When $k = 2$:
$z_3=8^{\frac{1}{3}}(\cos\frac{\pi+4\pi}{3}+i\sin\frac{\pi + 4\pi}{3})=2(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})=2\mathrm{cis}\frac{5\pi}{3}$

Step5: Differentiate $y=x^{4}-5$

Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ and $\frac{d}{dx}(c)=0$ (where $c$ is a constant), we have $\frac{dy}{dx}=\frac{d}{dx}(x^{4}-5)=\frac{d}{dx}(x^{4})-\frac{d}{dx}(5)=4x^{3}-0 = 4x^{3}$.

Step6: Differentiate $f(x)=\cos(5x)$

Using the chain - rule $\frac{d}{dx}(f(g(x)))=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = 5x$, then $f(u)=\cos u$. $f^\prime(u)=-\sin u$ and $g^\prime(x)=5$. So $f^\prime(x)=-\sin(5x)\cdot5=-5\sin(5x)$.

Answer:

• $\frac{7x}{x^{2}-7x + 10}=\frac{35/3}{x - 5}-\frac{14/3}{x - 2}$
• $z_1 = 2\mathrm{cis}\frac{\pi}{3}$, $z_2 = 2\mathrm{cis}\pi$, $z_3 = 2\mathrm{cis}\frac{5\pi}{3}$
• $\frac{dy}{dx}=4x^{3}$
• $f^\prime(x)=-5\sin(5x)$