# Step-by-step Solution

## Find the derivative of $\sin\left(2x\right)$

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### Videos

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

## Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(sin\left(2x\right)\right)$

Choose the solving method

1

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 $\sin\left(2x\right)$. Substituting $f(x+h)$ and $f(x)$ on the limit

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

Solve the product $2\left(x+h\right)$

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

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

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

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

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

Split the fraction $\frac{\sin\left(2x\right)\left(\cos\left(2h\right)-1\right)+\cos\left(2x\right)\sin\left(2h\right)}{h}$ in two fractions with common denominator $h$

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

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

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

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

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

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

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

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

$\sin\left(2x\right)\lim_{h\to0}\left(\frac{\cos\left(2h\right)-1}{h}\right)+2\cos\left(2x\right)$
10

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

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

$2\cos\left(2x\right)$
$\frac{d}{dx}\left(sin\left(2x\right)\right)$

### Main topic:

Differential calculus

### Time to solve it:

~ 0.36 s (SnapXam)