How do you differentiate #f(x)=csc(e^(x^3-x)) # using the chain rule?

Question

How do you differentiate #f(x)=csc(e^(x^3-x))  # using the chain rule?

Cameron Alexander · Accepted Answer

#f' (x)=(1-3x^2)(e^(x^3-x))*csc(e^(x^3-x)) *cot(e^(x^3-x))#

Christopher Agresta · Answer

undefined To differentiate $ f(x) = \csc(e^{x^3 - x}) $ using the chain rule, follow these steps:

1. Identify the outer function, which is $\csc(u)$, where $ u = e^{x^3 - x} $.
2. Differentiate the outer function with respect to its inner variable, $ u $, using the derivative of the cosecant function: $ \frac{d}{du} \csc(u) = -\csc(u) \cot(u) $.
3. Find the derivative of the inner function, $ u = e^{x^3 - x} $, using the chain rule.
4. Combine the results of steps 2 and 3 to obtain the derivative of the entire function $ f(x) $.

The derivative of $ f(x) $ with respect to $ x $ is:

$$ f'(x) = -\csc(e^{x^3 - x}) \cot(e^{x^3 - x}) \cdot \frac{d}{dx}(e^{x^3 - x}) $$

To find $ \frac{d}{dx}(e^{x^3 - x}) $, differentiate $ e^{x^3 - x} $ using the chain rule:

$$ \frac{d}{dx}(e^{x^3 - x}) = e^{x^3 - x} \cdot \frac{d}{dx}(x^3 - x) $$

$$ = e^{x^3 - x} \cdot (3x^2 - 1) $$

So, combining everything:

$$ f'(x) = -\csc(e^{x^3 - x}) \cot(e^{x^3 - x}) \cdot e^{x^3 - x} \cdot (3x^2 - 1) $$

$$ = - (3x^2 - 1) \csc(e^{x^3 - x}) \cot(e^{x^3 - x}) e^{x^3 - x} $$