# Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. y=  int((3t+(t)^(1/2))^(1/2)))dt, t=7...tanx)) Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. y=  int((3t+(t)^(1/2))^(1/2)))dt, t=7...tanx)) y' = ________________ y=int_7^(tan(x))sqrt(3t+sqrt(t)))dt

The first part of the theorem says,if,

F(x) = int_a^xf(t)dt then,

F'(x) = f(x)

In our case, we have tan x instead of x, so we can't apply this theorem straight away, so we need a substitution. we will use

u =tan(x)

so , (du)/(dx) = sec^2(x)

now...

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y=int_7^(tan(x))sqrt(3t+sqrt(t)))dt

The first part of the theorem says,if,

F(x) = int_a^xf(t)dt then,

F'(x) = f(x)

In our case, we have tan x instead of x, so we can't apply this theorem straight away, so we need a substitution. we will use

u =tan(x)

so , (du)/(dx) = sec^2(x)

now our integral becomes,

y=int_7^usqrt(3t+sqrt(t)))dt

but we need '0' on the lower limit.

y=int_0^usqrt(3t+sqrt(t)))dt - int_0^7sqrt(3t+sqrt(t))dt

so,
dy/du becomes,

(dy)/(du) = f(u) - f(7)

(dy)/(du) = sqrt(3u+sqrt(u)) - sqrt(3*7+sqrt(7))

(dy)/(du) = sqrt(3u+sqrt(u)) - 4.862689

But we know as per chain rule,

(dy)/(dx) = (dy)/(du) * (du)/(dx)

so we get,

(dy)/(dx) = (sqrt(3u+sqrt(u)) - 4.862689) * sec^2(x)

but u is tan(x)

(dy)/(dx) = (sqrt(3tan(x)+sqrt(tan(x))) - 4.862689) * sec^2(x)

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