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Find the derivative of $\cos\left(2x\right)$ using the definition

Step-by-step Solution

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Final Answer

$-2\sin\left(2x\right)$
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Step-by-step Solution

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1

Find the derivative of $\cos\left(2x\right)$ using the definition. Apply the definition of the derivative: $\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. The function $f(x)$ is the function we want to differentiate, which is $\cos\left(2x\right)$. Substituting $f(x+h)$ and $f(x)$ on the limit, we get

$\lim_{h\to0}\left(\frac{\cos\left(2\left(x+h\right)\right)-\cos\left(2x\right)}{h}\right)$
2

Multiply the single term $2$ by each term of the polynomial $\left(x+h\right)$

$\lim_{h\to0}\left(\frac{\cos\left(2x+2h\right)-\cos\left(2x\right)}{h}\right)$
3

Using the cosine of a sum formula: $\cos(\alpha\pm\beta)=\cos(\alpha)\cos(\beta)\mp\sin(\alpha)\sin(\beta)$, where angle $\alpha$ equals $2x$, and angle $\beta$ equals $2h$

$\lim_{h\to0}\left(\frac{\cos\left(2x\right)\cos\left(2h\right)-\sin\left(2x\right)\sin\left(2h\right)-\cos\left(2x\right)}{h}\right)$
4

Factoring by $\cos\left(2x\right)$

$\lim_{h\to0}\left(\frac{\cos\left(2x\right)\left(\cos\left(2h\right)-1\right)-\sin\left(2x\right)\sin\left(2h\right)}{h}\right)$
5

Expand the fraction $\frac{\cos\left(2x\right)\left(\cos\left(2h\right)-1\right)-\sin\left(2x\right)\sin\left(2h\right)}{h}$ into $2$ simpler fractions with common denominator $h$

$\lim_{h\to0}\left(\frac{\cos\left(2x\right)\left(\cos\left(2h\right)-1\right)}{h}+\frac{-\sin\left(2x\right)\sin\left(2h\right)}{h}\right)$
6

The limit of a sum of two or more functions is equal to the sum of the limits of each function: $\displaystyle\lim_{x\to c}(f(x)\pm g(x))=\lim_{x\to c}(f(x))\pm\lim_{x\to c}(g(x))$

$\lim_{h\to0}\left(\frac{\cos\left(2x\right)\left(\cos\left(2h\right)-1\right)}{h}\right)+\lim_{h\to0}\left(\frac{-\sin\left(2x\right)\sin\left(2h\right)}{h}\right)$
7

The limit of the product of a function and a constant is equal to the limit of the function, times the constant: $\displaystyle \lim_{t\to 0}{\left(at\right)}=a\cdot\lim_{t\to 0}{\left(t\right)}$

$\cos\left(2x\right)\lim_{h\to0}\left(\frac{\cos\left(2h\right)-1}{h}\right)+\lim_{h\to0}\left(\frac{-\sin\left(2x\right)\sin\left(2h\right)}{h}\right)$
8

The limit of the product of a function and a constant is equal to the limit of the function, times the constant: $\displaystyle \lim_{t\to 0}{\left(at\right)}=a\cdot\lim_{t\to 0}{\left(t\right)}$

$\cos\left(2x\right)\lim_{h\to0}\left(\frac{\cos\left(2h\right)-1}{h}\right)+\sin\left(2x\right)\lim_{h\to0}\left(\frac{-\sin\left(2h\right)}{h}\right)$
9

Knowing that $\displaystyle\lim_{h\to 0}{\left(\frac{\cos(h)-1}{h}\right)}=0$

$0\cos\left(2x\right)$
10

Any expression multiplied by $0$ is equal to $0$

0
11

Any expression multiplied by $0$ is equal to $0$

$0+\sin\left(2x\right)\lim_{h\to0}\left(\frac{-\sin\left(2h\right)}{h}\right)$
12

$x+0=x$, where $x$ is any expression

$\sin\left(2x\right)\lim_{h\to0}\left(\frac{-\sin\left(2h\right)}{h}\right)$
13

The limit of the product of a function and a constant is equal to the limit of the function, times the constant: $\displaystyle \lim_{t\to 0}{\left(at\right)}=a\cdot\lim_{t\to 0}{\left(t\right)}$

$-\sin\left(2x\right)\lim_{h\to0}\left(\frac{\sin\left(2h\right)}{h}\right)$
14

Apply the formula: $\lim_{h\to0}\left(\frac{\sin\left(nh\right)}{h}\right)$$=n$, where $n=2$

$-2\sin\left(2x\right)$

Final Answer

$-2\sin\left(2x\right)$

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Plotting: $-2\sin\left(2x\right)$

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7
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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Definition of Derivative

Resolution of derivatives using the definition of the derivative, which is the limit of difference quotients of real numbers.

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