How do you differentiate g(x) =x^2tanx using the product rule?

Feb 8, 2016

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

Explanation:

The product rule states that the derivative of $u v$, where $u$ and $v$ are functions of $x$, is $u ' v + u v '$. In other words, it's the derivative of one function times the other plus the derivative of the other function times the other.

We solve these problems by first finding the derivatives of each piece. In this case, we have $u = {x}^{2}$ and $v = \tan x$ (note that $v$ can be ${x}^{2}$ and $u$ can be $\tan x$, it really doesn't matter).
$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$
$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$

Now we substitute $u$ and $v$ and their derivatives into the product rule formula:
$g ' \left(x\right) = \left({x}^{2}\right) ' \left(\tan x\right) + \left({x}^{2}\right) \left(\tan x\right) '$
$g ' \left(x\right) = 2 x \tan x + {x}^{2} {\sec}^{2} x$

We could rewrite this in other ways, like $x \sec x \left(2 \sin x + \sec x\right)$, but they all mean the same thing. If your instructor prefers writing it using only sines and cosines, with the help of identities, you can express the result as $\frac{x \sin 2 x + {x}^{2}}{\cos} ^ 2 x$.

Feb 8, 2016

g'(x) = ${x}^{2} {\sec}^{2} x + 2 x \tan x$

Explanation:

using the product rule :

If g(x) = f(x).h(x) then g'(x) = f(x).h'(x) + h(x).f'(x)

here let f(x) $= {x}^{2} \textcolor{b l a c k}{\text{ and }} h \left(x\right) = \tan x$

hence g'(x)$= {x}^{2} \frac{d}{\mathrm{dx}} \left(\tan x\right) + \tan x \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

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