I'm struggling trying to find the second derivative of
f(x)=x^4*e^x
I've found f ' (x)= e^x*x^3(4+x) using the product rule.
I'm not very good with the dy/dx notations yet and have been using u, v, u' and v' when doing my equations, so f ' (x) = [(u' * v) + (v' * u)] / (v)^2
So:
u=x^4
u'=4x^3
v=e^x
v'=e^x
I understand that I will need to get more familiar with the dy/dx format, but I'm not quite there and am looking for a simple formal to calculate f '' (x) in terms of u and v.
Thank you!!
f(x)=x^4*e^x
I've found f ' (x)= e^x*x^3(4+x) using the product rule.
I'm not very good with the dy/dx notations yet and have been using u, v, u' and v' when doing my equations, so f ' (x) = [(u' * v) + (v' * u)] / (v)^2
So:
u=x^4
u'=4x^3
v=e^x
v'=e^x
I understand that I will need to get more familiar with the dy/dx format, but I'm not quite there and am looking for a simple formal to calculate f '' (x) in terms of u and v.
Thank you!!
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What you have for f' is correct, but you shouldn't have pulled out the factor and simplified. This is because it will be easier to find f" if you leave it:
f' = x^4*e^x + e^x*4x^3. Now find f" of each term separately.
Use the product rule on x^4*e^x,
and then use the product ruleon e^x*4x^3.
Then, you may want to pull out a common factor as this will be your final answer.
You seem to know what you are doing, so i think you will be fine from here.
email w/questions
p.s. [(u' * v) + (v' * u)] / (v)^2 is the quotient rule. oops!!
dy/dx means the derivative of the y function (f(x), with respect to x.
Basically, this simply means x' = 1
Soon you will do implicit differentiation with respect to x.
dy/dx using implicit differentiation means the derivative of X^2 = 2x,
while the derivative of y^2 is 2yy', because in dy/dx: y' = y', not 1
your teacher will explain when the time comes
Additional Info:
This stuff about dy/dx is for future use, not this problem.
Hope I didn't confuse?
f' = x^4*e^x + e^x*4x^3. Now find f" of each term separately.
Use the product rule on x^4*e^x,
and then use the product ruleon e^x*4x^3.
Then, you may want to pull out a common factor as this will be your final answer.
You seem to know what you are doing, so i think you will be fine from here.
email w/questions
p.s. [(u' * v) + (v' * u)] / (v)^2 is the quotient rule. oops!!
dy/dx means the derivative of the y function (f(x), with respect to x.
Basically, this simply means x' = 1
Soon you will do implicit differentiation with respect to x.
dy/dx using implicit differentiation means the derivative of X^2 = 2x,
while the derivative of y^2 is 2yy', because in dy/dx: y' = y', not 1
your teacher will explain when the time comes
Additional Info:
This stuff about dy/dx is for future use, not this problem.
Hope I didn't confuse?
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I would have rated you both best if I could lol I solved the second derivative using the ways you both suggested and found them both easy once I understood what to do. Thank you for your help!!!
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The second deriviattive is simply the deriviative of the deriviative. So in this case,
f''(x) is the deriviative of e^x x^3(4+x)
As you've written it, there are 3 things multiplied together (e^x, x^3 and 4+x), so you've got to use the product rule twice. So I think it's easier to rewrite f'(x) (which you did correctly by the way, well done!) as:
f'(x) = e^x (x^4 + 4x^3)
Now set u and v for doing this deriviative:
u = e^x
v = (x^4 + 4x^3)
And so:
u' = e^x
v' = (4x^3 + 12x^2)
giving:
f'''(x) = u'v + v'u = e^x (x^4 + 4x^3) + e^x (4x^3 + 12x^2)
= e^x (x^4 + 8x^3 + 12x^2)
f''(x) is the deriviative of e^x x^3(4+x)
As you've written it, there are 3 things multiplied together (e^x, x^3 and 4+x), so you've got to use the product rule twice. So I think it's easier to rewrite f'(x) (which you did correctly by the way, well done!) as:
f'(x) = e^x (x^4 + 4x^3)
Now set u and v for doing this deriviative:
u = e^x
v = (x^4 + 4x^3)
And so:
u' = e^x
v' = (4x^3 + 12x^2)
giving:
f'''(x) = u'v + v'u = e^x (x^4 + 4x^3) + e^x (4x^3 + 12x^2)
= e^x (x^4 + 8x^3 + 12x^2)