Question

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

Answer 1

`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.`

Spread the Joy of Maths.!

Answer 2

`F'(x) = 2/x`

Explanation:

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

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

Answer 3

`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`