How do you use the chain rule to differentiate #f(x)=37-sec^3(2x)#?

Question

Isabella Ackerman · Accepted Answer

#-6tan(2x)sec^3(2x)#

To differentiate this function, we can ignore the constant part, which is 37, because it will become zero after differentiation.

Let's focus on #-sec^3(2x)#. You might want to rewrite it as #-(sec(2x))^3# for better conceptualisation when applying the chain rule.

Then, we can further simplify it by substituting a variable for the secant part: #-u^3#

We get #-3u^2u'# from that. #u'# represents the derivative of #sec(2x)#, which is #2sec(2x)tan(2x)#, number '2' comes from the #2x# part as its derivative.

What we have: #u=sec(2x)# #u'=2sec(2x)tan(2x)#

Substitute them back to #-3u^2u'#, we have #-3sec^2(2x)*2sec(2x)tan(2x)#.

Finally, clear up the expression by combining like terms. #-6tan(2x)sec^3(2x)#

The method I used is called u-sub, it will come handier when you deal with the integrals. It helps you avoid losing yourself in the process when handling with long chains. Of course, if you can keep a neat scratch, the u-sub is not necessary for this type of chain rule functions.

Hope it helps.

Emilia Aguilar · Answer

undefined To differentiate the function $ f(x) = 37 - \sec^3(2x) $ using the chain rule, follow these steps:

1. **Identify the Outer and Inner Functions**: The outer function is $ u^3 $ where $ u = \sec(v) $, and the inner function $ v = 2x $. Thus, we have a composition of three functions: $ u^3 $, $ \sec(v) $, and $ 2x $.

2. **Differentiate the Outermost Layer**: Differentiate $ u^3 $ with respect to $ u $ to get $ 3u^2 $.

3. **Differentiate the Middle Layer**: Differentiate $ \sec(v) $ with respect to $ v $ to get $ \sec(v)\tan(v) $.

4. **Differentiate the Innermost Layer**: Differentiate $ 2x $ with respect to $ x $ to get $ 2 $.

5. **Apply the Chain Rule**: Multiply the derivatives obtained in steps 2, 3, and 4, in the order of outer to inner.

The derivative of $ f(x) = 37 - \sec^3(2x) $ with respect to $ x $ is found by applying the chain rule as follows:

$$ f'(x) = -\frac{d}{dx}[\sec^3(2x)] $$

$$ = -3\sec^2(2x) \cdot \frac{d}{dx}[\sec(2x)] $$

$$ = -3\sec^2(2x) \cdot (\sec(2x)\tan(2x)) \cdot \frac{d}{dx}[2x] $$

$$ = -3\sec^2(2x) \cdot \sec(2x)\tan(2x) \cdot 2 $$

$$ = -6\sec^3(2x)\tan(2x) $$

Therefore, the derivative of $ f(x) = 37 - \sec^3(2x) $ with respect to $ x $ is:

$$ f'(x) = -6\sec^3(2x)\tan(2x) $$