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strojnjashka [21]

2 years ago

10

Differentiate 2sin(5x-3)

1 answer:

anygoal [31]2 years ago

3 0

$f(x)=2\sin(5x-3)\\
f'(x)=2\cos(5x-3)\cdot5=10\cos(5x-3)$

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I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

5x + 3 = 6x – 1<br> Solve for the solution

Genrish500 [490]

Answer: x=4

Step-by-step explanation:

Subtract 6x from both sides.

5x+3−6x=6x−1−6x

−x+3=−1

Step 2: Subtract 3 from both sides.

−x+3−3=−1−3

−x=−4

Step 3: Divide both sides by -1.

−x/−1 = −4/−1

x=4

4 0

2 years ago

Read 2 more answers

What is 2×3(10)^-5 in standard form

Sindrei [870]

2×3(10)^{-5}=6×(10)^-5= $\frac{6}{10^5} =0.00006$

5 0

3 years ago

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

skelet666 [1.2K]

-5t2 + 40t = 0
5t(-t+ 8) = 0
5t = 0 or t = 8

7 0

2 years ago

Help me please (no links please)

stich3 [128]

a) This part is already complete I think..

b) This is a cuboid and lateral surface area of cuboid is: 2(lb +bh + hl)

= 2( 10 × 3 + 3 × 7 + 7 × 10)

= 2(30 + 21 + 70)

= 2 × 121 = 242 cm²

Now, the area of top & bottom: lb

= 2 × 10 × 3

= 60 cm²

Neglecting the top & bottom surface area of cuboid:

= 242 - 60

= 180 cm²

c) The total surface area us 242.. I have already done that part above...

__________________________

If i have done something wrong.. please lemme know :)

7 0

2 years ago