F(x)= intergal (bottom) to (top) as x goes to x^2

sec^2 t dt

d(tant)/dt=sec^2t dt

so the integral of sec^2 tdt = tant

That isn't one of the answer choices

Its more detailed I believe

I believe you know have to plug in what the integral goes to

please clarify what integral goes from i am not able to get it....is it from x to x^2

answer should be A)

Can you show how you got this answer? I need to know how to work the problem

the answer of integral is

(tant) from x to x^2

i e tanx^2-tanx

now we have to differentiate the function tanx^2- tanx

now we will get,

d/dt(tanx^2-tanx)=2xsec^2 x^2-sec^2x

=2x.secx^2.sec^2 -secx.secx

so answer is B sorry for mistake.

did u get????

Can you break down your derivative steps

ok.

d/dt(tanx^2-tanx)= d/dt(tan(x^2))-d/dt(tanx) but we know that d/dt(tanx)=sec^2 (x)

so = sec^2(x^2) .d/dt(x^2) -sec^2(x) ( use chain rule of derivative)

=2xsec^2 (x^2)-sec^2x

=2x.sec(x^2).sec(x^2) - secx.secx

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