If #f(x)= tan5 x # and #g(x) = 2x^2 -1 #, how do you differentiate #f(g(x)) # using the chain rule?

Question

If #f(x)= tan5 x # and #g(x)  = 2x^2 -1   #, how do you differentiate #f(g(x))  # using the chain rule?

Elijah Abram · Accepted Answer

#20xsec^2(10x^2-5)#

The chain rule states that

#d/dx[f(g(x))]=f'(g(x))*g'(x)#

First, find #f'(g(x))#.

#f'(x)=5sec^2(5x)#

Note that this required the chain rule as well.

#f'(g(x))=5sec^2(5(2x^2-1))=5sec^2(10x^2-5)#

Now, find #g'(x)#.

#g'(x)=4x#

Combine.

#d/dx[f(g(x))]=5sec^2(10x^2-5)*4x=20xsec^2(10x^2-5)#

Axel Alvarez · Answer

undefined To differentiate $ f(g(x)) $ using the chain rule, follow these steps:

1. Find $ f'(g(x)) $ by differentiating $ f(x) $ with respect to its inner function $ g(x) $.
2. Find $ g'(x) $.
3. Multiply $ f'(g(x)) $ and $ g'(x) $ together.

Given $ f(x) = \tan^5(x) $ and $ g(x) = 2x^2 - 1 $:

1. Find $ f'(g(x)) $:
$$ f'(x) = 5\tan^4(x)\sec^2(x) $$
$$ f'(g(x)) = 5\tan^4(2x^2 - 1)\sec^2(2x^2 - 1) $$

2. Find $ g'(x) $:
$$ g'(x) = 4x $$

3. Multiply $ f'(g(x)) $ and $ g'(x) $:
$$ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) $$
$$ \frac{d}{dx}[f(g(x))] = 5\tan^4(2x^2 - 1)\sec^2(2x^2 - 1) \cdot 4x $$

So, $ \frac{d}{dx}[f(g(x))] = 20x\tan^4(2x^2 - 1)\sec^2(2x^2 - 1) $.