# How to find first derivative of f(x)=2 sin (3x) + x?

Aug 7, 2015

$f ' \left(x\right) = 6 \cos \left(3 x\right) + 1$

#### Explanation:

Differentiate each term:

$\frac{d \left(x\right)}{\mathrm{dx}} = 1$

Using the chain rules for the second term we have:

$g \left(x\right) = h \left(k \left(x\right)\right) \implies g ' \left(x\right) = k ' \left(x\right) h ' \left(k \left(x\right)\right)$

With:
$h \left(u\right) = 2 \sin \left(u\right) \implies h ' \left(u\right) = 2 \cos \left(u\right)$
$k \left(x\right) = 3 x \implies k ' \left(x\right) = 3$
$g \left(x\right) = 2 \sin \left(3 x\right) \implies g ' \left(x\right) = 6 \cos \left(3 x\right)$

Together we have:
$f ' \left(x\right) = 6 \cos \left(3 x\right) + 1$

Aug 7, 2015

We are asked to find the derivative of $f \left(x\right) = 2 \sin \left(3 x\right) + x$ using the definition: $f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$.

#### Explanation:

We need to evaluate:

${\lim}_{h \rightarrow 0} \frac{{\overbrace{\left[2 \sin \left(3 \left(x + h\right)\right) + \left(x + h\right)\right]}}^{f \left(x + h\right)} - {\overbrace{\left[2 \sin \left(3 x\right) + x\right]}}^{f} \left(x\right)}{h}$.

This will be cumbersome. To make it look less complicated, let's split the expression into two simpler parts. We'll take the trigonometric part and the linear part separately.

${\lim}_{h \rightarrow 0} \frac{2 \sin \left(3 \left(x + h\right)\right) - 2 \sin 3 x}{h} + {\lim}_{h \rightarrow 0} \frac{\left(x + h\right) - x}{h}$

I will assume that you can show that the second limit is $1$. The more challenging limit is the limit involving trigonometric functions.

${\lim}_{h \rightarrow 0} \frac{2 \sin \left(3 \left(x + h\right)\right) - 2 \sin 3 x}{h} = 2 {\lim}_{h \rightarrow 0} \frac{\sin \left(3 x + 3 h\right) - \sin 3 x}{h}$

$= 2 {\lim}_{h \rightarrow 0} \frac{{\overbrace{\left(\sin 3 x \cos 3 h + \cos 3 x \sin 3 h\right)}}^{\sin} \left(3 x + 3 h\right) - \sin 3 x}{h}$

$= 2 {\lim}_{h \rightarrow 0} \frac{\sin 3 x \cos 3 x - \sin 3 x + \cos 3 x \sin 3 x}{h}$

$= 2 {\lim}_{h \rightarrow 0} \left(\frac{\sin 3 x \left(\cos 3 h - 1\right)}{h} + \frac{\cos 3 x \sin 3 h}{h}\right)$

$= 2 {\lim}_{h \rightarrow 0} \left(\sin 3 x \frac{\cos 3 h - 1}{h} + \cos 3 x \frac{\sin 3 h}{h}\right)$

$= 2 \left[{\lim}_{h \rightarrow 0} \sin 3 x {\lim}_{h \rightarrow 0} \frac{\cos 3 h - 1}{h} + {\lim}_{h \rightarrow 0} \cos 3 x {\lim}_{h \rightarrow 0} \frac{\sin 3 h}{h}\right]$

$= 2 \left[\left({\lim}_{h \rightarrow 0} \sin 3 x\right) \left(3 {\lim}_{h \rightarrow 0} \frac{\cos 3 h - 1}{3 h}\right) + \left({\lim}_{h \rightarrow 0} \cos 3 x\right) \left(3 {\lim}_{h \rightarrow 0} \frac{\sin 3 h}{3 h}\right)\right]$

$= 2 \left[\left(\sin 3 x\right) \left(3 \cdot 0\right) + \left(\cos 3 x\right) \left(3 \cdot 1\right)\right]$

$= 2 \left(3 \cos 3 x\right) = 6 \cos \left(3 x\right)$

So, when we put the two pieces together, we get:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\left[2 \sin \left(3 \left(x + h\right)\right) + \left(x + h\right)\right] - \left[2 \sin \left(3 x\right) + x\right]}{h}$

$= {\lim}_{h \rightarrow 0} \frac{2 \sin \left(3 \left(x + h\right)\right) - 2 \sin 3 x}{h} + {\lim}_{h \rightarrow 0} \frac{\left(x + h\right) - x}{h}$

$= 6 \cos \left(3 x\right) + 1$