How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #h(x) = int sqrt(6 + r^3) dx# from 7 to #x^2#?

Answer 1

We need FTC 1 and the chain rule.

#h(x) = int_7^(x^2) sqrt(6 + r^3) dx# does not quite fit the form:
#g(x) = int_a^b f(t) dt# whose derivative is #f(x)#

But we can use the chain rule here;

The #h(x)# can be written as
#h(u) = int_7^u sqrt(6 + r^3) dx# where #u = x^2#

So

#d/dx(h) = (dh)/(du)* (du)/dx#
# = sqrt(6 + u^3) (du)/dx#
# = sqrt(6 + (x^2)^3)*(2x)#
# = 2xsqrt(6 + x^6)#

Summary

For #h(x) = int_a^(g(x)) f(t) dt# #" "#(With the right kind of #f# and domain of #h#)

We have

#h'(x) = f(g(x))*g'(x)# #" "# (Which is one way of writing the chain rule.)
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Answer 2

To find the derivative of the function ( h(x) = \int_{7}^{x^2} \sqrt{6 + r^3} , dr ) using Part 1 of the Fundamental Theorem of Calculus, you need to evaluate the integrand function at the upper limit of integration, which is ( x^2 ), and then multiply by the derivative of the upper limit with respect to ( x ), which is ( 2x ). Therefore, the derivative of the function ( h(x) ) is ( h'(x) = \sqrt{6 + (x^2)^3} \cdot 2x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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