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

Question

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

Christopher Allers · Accepted Answer

# (sec ( e^ (sqrt(x))) * tan ( e^ (sqrt(x))) * e^ (sqrt(x))) /(2(sqrt(x)) # # d/dx [ sec ( - e^(sqrt(x))) ] # # sec ( e^ (sqrt(x))) * tan ( e^ (sqrt(x))) * d/dx [ ( e^ (sqrt(x))) ] # # sec ( e^ (sqrt(x))) * tan ( e^ (sqrt(x))) * e^ (sqrt(x)) * d/dx[ sqrt(x) ] # # sec ( e^ (sqrt(x))) * tan ( e^ (sqrt(x))) * e^ (sqrt(x)) * 1/(2(sqrt(x)) # # (sec ( e^ (sqrt(x))) * tan ( e^ (sqrt(x))) * e^ (sqrt(x))) /(2(sqrt(x)) #

Luke Alvarez · Answer

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

1. Identify the outer function: $ \sec(u) $, where $ u = -e^{\sqrt{x}} $.
2. Find the derivative of the outer function: $ \sec(u) $ has the derivative $ \sec(u) \tan(u) $.
3. Identify the inner function: $ u = -e^{\sqrt{x}} $.
4. Find the derivative of the inner function: $ \frac{du}{dx} = \frac{d}{dx}(-e^{\sqrt{x}}) $.
5. Apply the chain rule: $ \frac{d}{dx} \sec(-e^{\sqrt{x}}) = \sec(-e^{\sqrt{x}}) \tan(-e^{\sqrt{x}}) \cdot \frac{d}{dx}(-e^{\sqrt{x}}) $.

So, the differentiation of $ f(x) = \sec(-e^{\sqrt{x}}) $ using the chain rule is $ f'(x) = \sec(-e^{\sqrt{x}}) \tan(-e^{\sqrt{x}}) \cdot \frac{d}{dx}(-e^{\sqrt{x}}) $.