How do you find #F'(x)# given #F(x)=int 1/t dt# from #[1,x^2]#?

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Steve M Share
Jan 31, 2017

Answer:

# F'(x) = 2/x #

Explanation:

If asked to find the derivative of an integral using the fundamental theorem of Calculus, you should not evaluate the integral

The Fundamental Theorem of Calculus tells us that:

# d/dx \ int_1^x \ 1/t \ dt = 1/x #

(ie the derivative of an integral gives us the original function back). We are asked to find (notice the upper bound as changed from #x# to #x^2#)

# F'(x) = d/dx int_1^(x^2) \ 1/t \ dt #

Using the chain rule we can rewrite as:

# F'(x) = (d(x^2))/dx d/(d(x^2)) int_1^(x^2) \ 1/t \ dt #

Now, #(d(x^2))/dx = 2x #, And, using the first result from the FTOC:

#d/(d(x^2)) int_1^(x^2) \ 1/t \ dt # = 1/x^2

Hence combining these trivial results we get:

# \ \ \ \ \ F'(x) = 2x * 1/x^2 #
# :. F'(x) = 2/x #

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Jan 31, 2017

Answer:

#F'(x) = 2/x#

Explanation:

#F(x) = int_0^(x^2) (dt)/t = ln(x^2) -1#

#F'(x) = (2x)/x^2 = 2/x#

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Jan 31, 2017

Answer:

#F'(x)=2/x, xne0.#

Explanation:

#F(x)=int_1^(x^2) 1/t dt=[ln|t|]_1^(x^2)=lnx^2-ln1=2ln|x|#.

#"So, "F(x)=2lnx, if x>0;#

#=2ln(-x), if x<0.#

#"Accordingly, "F'(x)=2/x, if x>0; and,#

#F'(x)=(2/-x)(-x)'=(2/-x)(-1)=2/x, if x<0.#

#"Therefore, in either of the case, "F'(x)=2/x, xne0.#

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