# How do you differentiate f(x)=xtan3x+x^3tanx using the product rule?

Dec 15, 2015

Find both derivatives separately, each using the product rule.

$\frac{d}{\mathrm{dx}} \left[x \tan 3 x\right] = \tan 3 x \frac{d}{\mathrm{dx}} \left[x\right] + x \frac{d}{\mathrm{dx}} \left[\tan 3 x\right]$

$\frac{d}{\mathrm{dx}} \left[x\right] = 1$

$\frac{d}{\mathrm{dx}} \left[\tan 3 x\right] = {\sec}^{2} 3 x \cdot \frac{d}{\mathrm{dx}} \left[3 x\right] = 3 {\sec}^{2} 3 x$

Thus,

$\frac{d}{\mathrm{dx}} \left[x \tan 3 x\right] = \tan 3 x + 3 x {\sec}^{2} 3 x$

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$\frac{d}{\mathrm{dx}} \left[{x}^{3} \tan x\right] = \tan x \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] + {x}^{3} \frac{d}{\mathrm{dx}} \left[\tan x\right]$

$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] = 3 {x}^{2}$

$\frac{d}{\mathrm{dx}} \left[\tan x\right] = {\sec}^{2} x$

Thus,

$\frac{d}{\mathrm{dx}} \left[{x}^{3} \tan x\right] = 3 {x}^{2} \tan x + {x}^{3} {\sec}^{2} x$

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Add the two to find that

$f ' \left(x\right) = \tan 3 x + 3 x {\sec}^{2} 3 x + 3 {x}^{2} \tan x + {x}^{3} {\sec}^{2} x$