3x - y = 9 Slope-intercept??

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

Using pascal triangle; expand the following: i. (a+b)4 ii. (x–1 )3 iii. (a–b)6

Murrr4er [49]

Answer:

Every number of years 365 and I have been trying to

6 0

3 years ago

Just need someone to check my answer

mixas84 [53]

The correct answer is 216x⁶y⁵.

Explanation:
The first thing we do is raise the last monomial to the third power.

(4xy)(2x²y)(3xy)³
=(4xy)(2x²y)(3³x³y³)
=4xy(2x²y)(27x³y³).

Now we can multiply the first two monomials. When we multiply powers with the same base, we add the exponents:
8x³y²(27x³y³).

We multiply these last two monomials, again adding the exponents:
216x⁶y⁵.

3 0

3 years ago

Read 2 more answers

A 13-inch candle is lit and steadily burns until it is burned out.

brilliants [131]

Answer:

a. The remaining length of the candle varies from 10 inches to 5.5 inches

b. Remaining Length = 13 - b

Step-by-step explanation:

b.

The remaining (unburnt) length of the candle must be equal to the difference between total length of the candle and the burnt length of the candle. So, the expression for the remaining length can be written as:

Remaining Length = Total Length - Burnt Length

where,

Total length = 13 inch

Burnt Length = b

Therefore,

Remaining Length = 13 - b

b.

Now, we use the same expression for:

burned length = 3 inches

Therefore,

Remaining Length = 13 inches - 3 inches

Remaining Length = 10 inches

Now, for

burned length = 7.5 inches

Therefore,

Remaining Length = 13 inches - 7.5 inches

Remaining Length = 5.5 inches

Hence,

The remaining length of the candle varies from 10 inches to 5.5 inches

4 0

2 years ago

Factorise (l+m)² - (l- m)²

Svetach [21]

Answer:

use the formula of (a+b)^2

check out the picture

Step-by-step explanation:

4 0

2 years ago