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

Question

Elijah Abram · Accepted Answer

Use the product rule to independently find each of the two derivatives.

#d/dx[xtan3x]=tan3xd/dx[x]+xd/dx[tan3x]#

#d/dx[x]=1#

#d/dx[tan3x]=sec^2 3x*d/dx[3x]=3sec^2 3x#

Thus,

#d/dx[xtan3x]=tan3x+3xsec^2 3x#

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#d/dx[x^3tanx]=tanxd/dx[x^3]+x^3d/dx[tanx]#

#d/dx[x^3]=3x^2#

#d/dx[tanx]=sec^2x#

Thus,

#d/dx[x^3tanx]=3x^2tanx+x^3sec^2x#

——————————

To find that, add the two.

#f'(x)=tan3x+3xsec^2 3x+3x^2tanx+x^3sec^2x#

Paisley Johnson · Answer

undefined To differentiate the function $ f(x) = x\tan(3x) + x^3\tan(x) $ using the product rule, we first identify two functions: $ u(x) = x $ and $ v(x) = \tan(3x) + x^3\tan(x) $. Then, we differentiate each function with respect to $ x $ to find $ u'(x) $ and $ v'(x) $. Finally, we apply the product rule formula: $$ (uv)' = u'v + uv'. $$

The derivative $ u'(x) $ is simply $ 1 $, as the derivative of $ x $ with respect to $ x $ is $ 1 $.

For $ v'(x) $, we use the sum rule and the chain rule: $$ v'(x) = \frac{d}{dx}[\tan(3x)] + \frac{d}{dx}[x^3\tan(x)]. $$

Applying the chain rule, the derivative of $ \tan(3x) $ with respect to $ x $ is $ 3\sec^2(3x) $, and the derivative of $ x^3\tan(x) $ with respect to $ x $ involves both the product rule and the chain rule. We differentiate $ x^3 $ to get $ 3x^2 $ and then apply the product rule to $ x^2\tan(x) $, resulting in $ x^2\sec^2(x) + 2x\tan(x) $.

So, $ v'(x) = 3\sec^2(3x) + x^2\sec^2(x) + 2x\tan(x) $.

Finally, applying the product rule: $$ f'(x) = u'v + uv' = 1\cdot(\tan(3x) + x^3\tan(x)) + x\cdot(3\sec^2(3x) + x^2\sec^2(x) + 2x\tan(x)). $$

This simplifies to: $$ f'(x) = \tan(3x) + x^3\tan(x) + 3x\sec^2(3x) + x^3\sec^2(x) + 2x^2\tan(x). $$