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Mathematics 15 Online
OpenStudy (anonymous):

find the derivative of y with respect to x, t or theta. (a) e^7 - 10x (b) 8xe^x - 8e^x (c) y= (x^2 -2x+4)e^x (d) y= sin e^theta^4 please show the steps. thank you

OpenStudy (anonymous):

what is the derivative for a constant?

OpenStudy (anonymous):

zero?

OpenStudy (anonymous):

yeah what about -10x?

OpenStudy (anonymous):

is it -10?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

the answer to a is -10e^7-10x

OpenStudy (anonymous):

is that the correct answer or your?

OpenStudy (anonymous):

thats the correct answer but i just dont know how to get it

OpenStudy (anonymous):

(a) should be -10

OpenStudy (anonymous):

I think your missing some information

OpenStudy (anonymous):

is it y = e^7 - 10x?

OpenStudy (anonymous):

or e^(7-10x)?

OpenStudy (anonymous):

i think we r suppose to use the u substitution method. no the answer is -10e^7-10x

OpenStudy (anonymous):

what is the derivative of e^x?

OpenStudy (anonymous):

Are you familiar with the power rule and that any derivative of a constant will equal zero? \[\frac{d}{dx}(x^{n})=nx^{x-1}\]

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

the first one should be e^(7 - 10x)

OpenStudy (anonymous):

do you know the derivative of e^x?

OpenStudy (anonymous):

no is it x?

OpenStudy (anonymous):

no it is e^x

OpenStudy (anonymous):

http://www.math.com/tables/derivatives/tableof.htm

OpenStudy (anonymous):

do you know the chain rule?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

okay so what is the first one?

OpenStudy (anonymous):

You can pretend its e^u and u = 7-10x

OpenStudy (anonymous):

im not sure can u explain?

OpenStudy (anonymous):

so what is the derivative of e^x?

OpenStudy (anonymous):

e^x

OpenStudy (anonymous):

right what about e^u?

OpenStudy (anonymous):

e^u which is e^7-10x

OpenStudy (anonymous):

right now you have to use the chain rule by taking du/dx and multiply to e^u

OpenStudy (anonymous):

so du/dx * e^u is the answer.

OpenStudy (anonymous):

ok i think i got it! :)

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