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Mathematics 10 Online
OpenStudy (anonymous):

Find the derivative of the function. r(x) = (e^4x(squared)^6

OpenStudy (anonymous):

Don't take this as fact. You should double check me. Here I go

OpenStudy (anonymous):

\[6(e^{4x^2})^5(e^{4x^2})8x=48x(e^{4x^2})^6\]

OpenStudy (anonymous):

thats what i got instead i had it squared by 5 not 6

OpenStudy (anonymous):

I brought the power down then took the derivative of the inside, of which I had do the chain rule again.

OpenStudy (anonymous):

I combined the ^5 and ^1 with the same base to get ^6. But this is the first time I've done this exact type of problem. You very well could be right.

OpenStudy (anonymous):

oh you were right. thanks

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

can you help me with one more question, similiar to this?

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

\[e ^{2x^2-7x+2/x}\]

OpenStudy (anonymous):

thats a tough one check me on this first. The derivative of 1/x is ln(x) right?

OpenStudy (anonymous):

i think its this f ' (x) = -1 x ^ -2

OpenStudy (anonymous):

You're right integral of 1/x is ln(x) and derivative of ln(x) is 1/x. Ok. Think you have to bring x to the top so x^-1 then use the product rule. I'll try it

OpenStudy (anonymous):

is everything over x or just the 2

OpenStudy (anonymous):

just 2

OpenStudy (anonymous):

I was assuming everything was over x. This makes it simpler. You should still make it 2x^-1, but we don't have to use the product rule anymore

OpenStudy (anonymous):

\[e^{2x^2-7x+2/x}(4x-7-2/x^2)\]

OpenStudy (anonymous):

so everything is over x?

OpenStudy (anonymous):

No, I thought the original problem was over x, but you said it was only the 2. So my answer above the 2 is only over x^2

OpenStudy (anonymous):

oh ok, hm i guess i was wrong because it was incorrect for my hwk.

OpenStudy (anonymous):

im not sure how to do this type of problem

OpenStudy (anonymous):

\[e^{(\frac{ 2x^2-7x+2 }{ x })}\]Is that the problem

OpenStudy (anonymous):

no its labeled like it is a few convos up. when u gave me the answer

OpenStudy (anonymous):

Hmm. I wouldn't do anything different with my answer a few posts above. You should get a second opinion on this one I guess.

OpenStudy (anonymous):

hm ok. this is a tough one.

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