Differentiation using chain/product rule

  • Thread starter Baartzy89
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  • #1
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Hi,

Just a question on an example in a maths textbook. See attached image for question below.

So, I understand that if you set u=sin(x) and v=e^-cos(x)
f'(x)=u'.v + u.v'

But I'm stuck looking at e^-cos(x), could it also be classified e^(w)?

Also, the second step in differentiating the above equation seems to only differentiate the -cos(x) of the e^-cos(x)...is this because the derivative of e is simply e?
 

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Answers and Replies

  • #2
chiro
Science Advisor
4,790
132
Hey Baartzy89 and welcome to the forums.

You have the right idea about the exponential: we have two properties.

The first is that d/dx e^(x) = e^(x) and the chain rule which says d/dx f(g(x)) = g'(x)f'(g(x)), so you need to consider g(x) and the derivative of f(g(x)).
 

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