Derivatives with e^x - Am I doing this right?

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The function is f with f(x) = x[sup:2ow0yj1m]2[/sup:2ow0yj1m] + e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]
Find the first two derivatives.

Here is how I've tried this: I take the derivative of the term x[sup:2ow0yj1m]2[/sup:2ow0yj1m] and then add the derivative of e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m] by using the chain rule.
I understand the chain rule as follows: when f(x) = v(u(x)) then f'(x) = v'(x) u'(v(x)). And if f(x) = e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m], then accordingly,
f'(x) = -2 e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]. So my answers look as follows:

f(x) = x[sup:2ow0yj1m]2[/sup:2ow0yj1m] + e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]
f'(x) = 2x - 2e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]
f''(x) = 2 + 4e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]

Did I solve this correctly? The solutions are marked in my book as being:

f(x) = x[sup:2ow0yj1m]2[/sup:2ow0yj1m] + e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]
f'(x) = (2x - x[sup:2ow0yj1m]2[/sup:2ow0yj1m]) e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]
f''(x) = (x[sup:2ow0yj1m]2[/sup:2ow0yj1m] - 4x + 2) e[sup:2ow0yj1m]-2x[/sup:2ow0yj1m]

But I have no idea how these answers were achieved, and I can't see any relationship between mine and these actual solutions. What is going on here? And if I have done this terribly wrong, could you help me get on the right track?

 

[galactus]

The function is f with f(x) = x[sup:8iroanyw]2[/sup:8iroanyw] + e[sup:8iroanyw]-2x[/sup:8iroanyw]
Find the first two derivatives.

Here is how I've tried this: I take the derivative of the term x[sup:8iroanyw]2[/sup:8iroanyw] and then add the derivative of e[sup:8iroanyw]-2x[/sup:8iroanyw] by using the chain rule.
I understand the chain rule as follows: when f(x) = v(u(x)) then f'(x) = v'(x) u'(v(x)). And if f(x) = e[sup:8iroanyw]-2x[/sup:8iroanyw], then accordingly,
f'(x) = -2 e[sup:8iroanyw]-2x[/sup:8iroanyw]. So my answers look as follows:

f(x) = x[sup:8iroanyw]2[/sup:8iroanyw] + e[sup:8iroanyw]-2x[/sup:8iroanyw]
f'(x) = 2x - 2e[sup:8iroanyw]-2x[/sup:8iroanyw]
f''(x) = 2 + 4e[sup:8iroanyw]-2x[/sup:8iroanyw]

These are correct.

Did I solve this correctly?

Yes

The solutions are marked in my book as being:

f(x) = x[sup:8iroanyw]2[/sup:8iroanyw] + e[sup:8iroanyw]-2x[/sup:8iroanyw]
f'(x) = (2x - x[sup:8iroanyw]2[/sup:8iroanyw]) e[sup:8iroanyw]-2x[/sup:8iroanyw]
f''(x) = (x[sup:8iroanyw]2[/sup:8iroanyw] - 4x + 2) e[sup:8iroanyw]-2x[/sup:8iroanyw]

Is this what the book has as the derivatives to your problem?. If so, they are incorrect. You have the correct derivatives.

 

[Deleted member 4993]

The function is f with f(x) = x[sup:15us1o2p]2[/sup:15us1o2p] + e[sup:15us1o2p]-2x[/sup:15us1o2p]
Find the first two derivatives.

Here is how I've tried this: I take the derivative of the term x[sup:15us1o2p]2[/sup:15us1o2p] and then add the derivative of e[sup:15us1o2p]-2x[/sup:15us1o2p] by using the chain rule.
I understand the chain rule as follows: when f(x) = v(u(x)) then f'(x) = v'(x) u'(v(x)). And if f(x) = e[sup:15us1o2p]-2x[/sup:15us1o2p], then accordingly,
f'(x) = -2 e[sup:15us1o2p]-2x[/sup:15us1o2p]. So my answers look as follows:

f(x) = x[sup:15us1o2p]2[/sup:15us1o2p] + e[sup:15us1o2p]-2x[/sup:15us1o2p]
f'(x) = 2x - 2e[sup:15us1o2p]-2x[/sup:15us1o2p]
f''(x) = 2 + 4e[sup:15us1o2p]-2x[/sup:15us1o2p]

Did I solve this correctly? The solutions are marked in my book as being:

f(x) = x[sup:15us1o2p]2[/sup:15us1o2p] + e[sup:15us1o2p]-2x[/sup:15us1o2p]
f'(x) = (2x - x[sup:15us1o2p]2[/sup:15us1o2p]) e[sup:15us1o2p]-2x[/sup:15us1o2p]
f''(x) = (x[sup:15us1o2p]2[/sup:15us1o2p] - 4x + 2) e[sup:15us1o2p]-2x[/sup:15us1o2p]

But I have no idea how these answers were achieved, and I can't see any relationship between mine and these actual solutions. What is going on here? And if I have done this terribly wrong, could you help me get on the right track?

Are you sure that the given function was not as follows:

f(x) = x[sup:15us1o2p]2[/sup:15us1o2p] [sub:15us1o2p]*[/sub:15us1o2p] e[sup:15us1o2p]-x[/sup:15us1o2p]

thenthe answers are correct (sort of)!

then the answers could be

f'(x) = 2xe[sup:15us1o2p]-x[/sup:15us1o2p] - x[sup:15us1o2p]2[/sup:15us1o2p]e[sup:15us1o2p]-x[/sup:15us1o2p] = (2x - x[sup:15us1o2p]2[/sup:15us1o2p])e[sup:15us1o2p]-x[/sup:15us1o2p]

.

 

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Oh good, I am glad that it's not all wrong...

S. Khan is right, the correct function f for the answers in the book is:
f(x) = x[sup:3cf35w2v]2[/sup:3cf35w2v] * e[sup:3cf35w2v]-x[/sup:3cf35w2v]

The book must have slipped through a FEW stages of editing. There are many spelling mistakes, and sometimes this same sort of logical displacement such as writing the solutions to a problem which doesn't match. At least there are very few mathematical mistakes in it.
Thank you for your help!