given t(x)=2sec(x)-4cos(x), what is t'(x)?

here we want the derivative of the function

don't let the notation scare you t(x) is basically the same as f(x)

Alright so when finding the derivitive what do i do first?

do each term separate

first ind the derivative of sec(x)

that is something that is usually just memorized.. is probably in a table in the front or back of your book

secxtanx

yep =)

then what?

and the second term?

-sinx

so so far we have 2secxtanx+4cosx?

now what?

perfect thats it!

some people might dx in each term to show the chain rule.. but its usually not necessary

how would i do that?

well have you learned the chain rule yet?

Yes...but i dont understand it.

ok lets suppose we have this function: [f(x)= \sin (5x^2)\]

\[f(x)=\sin(5x^2)\]

Okay.

t'(x)=2sec(x)tan(x)+4sin(x)

so its not cos(x) its sin(x)

first you take the derivative of the outside \[\cos(5x^2)\] then you times it by the derivative of whats on the inside \[\cos(5x^2)(10x)\]

since derivative of -4cos(x) is -4(-sin(x))=4sin(x)

ahh yes your right myiniay missed that

I got it! Finally...its only been 3 weeks.

yayy

its really not that bad once you get it.. some people just try to make over complicate it just remember this: "take the derivative of what on the outside and times it by the derivative of whats on the inside"

its just hard to remmeber that the outside includes the inside.

yeaa dont touch the inside at first

gotta do the inside separate

want to try another one? that uses the chain rule?

sure! but other people keep answering them before i can even read the whole problem!!!

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