How do you use the second fundamental theorem of Calculus to find the derivative of given #int cos(t) / t dt # from #[3, sqrtx]#?

Answer 1

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = cossqrt(x)/(2x) #

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

# d/dx \ int_a^x \ f(t) \ dt = f(x) # for any constant #a#

(ie the derivative of an integral gives us the original function back).

We are asked to find:

# d/dx \ = int_3^(sqrt(x)) \ cost/t \ dt# ..... [A}

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

# u=sqrt(x) => (du)/dx = 1/2x^(-1/2) = 1/(2sqrt(x)) #

The substituting into the integral [A], and applying the chain rule, we get:

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = d/dx \ int_3^u \ cost/t \ dt #
# " " = (du)/dx*d/(du) \ int_3^u \ cost/t \ dt #
# " " = 1/(2sqrt(x)) \ d/(du) \ int_3^u \ cost/t \ dt #

Now that the integral's derivative is in the proper format for application of the FTOC, we obtain:

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = 1/(2sqrt(x)) \ cosu/u #

And going back to the original substitution, we obtain:

# d/dx \ int_3^(sqrt(x)) \ cost/t \ dt = 1/(2sqrt(x)) \ cossqrt(x)/sqrt(x) # # " " = cossqrt(x)/(2x) #
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Answer 2

To use the second fundamental theorem of calculus to find the derivative of ( \int_{3}^{\sqrt{x}} \frac{\cos(t)}{t} , dt ), you first need to evaluate the integral function. Then, differentiate the result with respect to ( x ). The second fundamental theorem of calculus states that if ( F(x) = \int_{a(x)}^{b(x)} f(t) , dt ), where ( a(x) ) and ( b(x) ) are functions, then ( F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) ). Apply this theorem to differentiate the integral function with respect to ( x ).

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Answer 3

To use the second fundamental theorem of calculus to find the derivative of ( \int_{3}^{\sqrt{x}} \frac{\cos(t)}{t} , dt ), we first need to define a function ( F(x) ) as follows:

[ F(x) = \int_{3}^{\sqrt{x}} \frac{\cos(t)}{t} , dt ]

Then, to find ( F'(x) ), we differentiate ( F(x) ) with respect to ( x ) using the second fundamental theorem of calculus, which states that if ( F(x) = \int_{a}^{g(x)} f(t) , dt ), then ( F'(x) = f(g(x)) \cdot g'(x) ):

[ F'(x) = \frac{\cos(\sqrt{x})}{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - \frac{\cos(3)}{3} \cdot 0 ] [ F'(x) = \frac{\cos(\sqrt{x})}{2x} ]

So, the derivative of the given integral with respect to ( x ) is ( \frac{\cos(\sqrt{x})}{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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