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mariarad [96]
3 years ago
7

Sara worked 40 hours per week

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
5 0
What is the question?
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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
Pls help<br> Find the measurements of those
ch4aika [34]

Answer:

i think C

im not really sure

8 0
2 years ago
JK and LM are perpendicular diameters of a circle. They are each 12 inches long. What is the approximate length of chord LK?
MA_775_DIABLO [31]

ANSWER:

I think the approximate length og chord Lk is around 8.5 inches long

3 0
3 years ago
Two functions are combined resulting in the function j(x) = –3x2 + 5.
labwork [276]

Answer:

two quadratic functions combined by addition

a linear function and a quadratic function combined by addition

two linear functions combined by multiplication

Step-by-step explanation:

-3x² + 5

(-4x² + 4) + (x² + 1)

-3x² + 5

(-3x² - x) + (x + 5)

-3x² + 5

(sqrt(5) - sqrt(3)x)(sqrt(5) + sqrt(3)x)

7 0
3 years ago
Type the correct answer in the box. Use numerals instead of words.
Nitella [24]

Answer:

35cm²

Step-by-step explanation:

Area = trapezium + triangle

Trapezium = ½×(8+4)×2 = 12

Triangle = ½×8×5.75 = 23

Area = 12+23 = 35 cm²

4 0
3 years ago
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