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

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# F'(x) = 2/x #

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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

# 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/(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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