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

Jan 31, 2017

$F ' \left(x\right) = \frac{2}{x} , x \ne 0.$

#### Explanation:

$F \left(x\right) = {\int}_{1}^{{x}^{2}} \frac{1}{t} \mathrm{dt} = {\left[\ln | t |\right]}_{1}^{{x}^{2}} = \ln {x}^{2} - \ln 1 = 2 \ln | x |$.

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

$= 2 \ln \left(- x\right) , \mathmr{if} x < 0.$

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

$F ' \left(x\right) = \left(\frac{2}{-} x\right) \left(- x\right) ' = \left(\frac{2}{-} x\right) \left(- 1\right) = \frac{2}{x} , \mathmr{if} x < 0.$

$\text{Therefore, in either of the case, } F ' \left(x\right) = \frac{2}{x} , x \ne 0.$

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Jan 31, 2017

$F ' \left(x\right) = \frac{2}{x}$

#### Explanation:

$F \left(x\right) = {\int}_{0}^{{x}^{2}} \frac{\mathrm{dt}}{t} = \ln \left({x}^{2}\right) - 1$

$F ' \left(x\right) = \frac{2 x}{x} ^ 2 = \frac{2}{x}$

Jan 31, 2017

$F ' \left(x\right) = \frac{2}{x}$

#### 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:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{x} \setminus \frac{1}{t} \setminus \mathrm{dt} = \frac{1}{x}$

(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 ${x}^{2}$)

$F ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{1}^{{x}^{2}} \setminus \frac{1}{t} \setminus \mathrm{dt}$

Using the chain rule we can rewrite as:

$F ' \left(x\right) = \frac{d \left({x}^{2}\right)}{\mathrm{dx}} \frac{d}{d \left({x}^{2}\right)} {\int}_{1}^{{x}^{2}} \setminus \frac{1}{t} \setminus \mathrm{dt}$

Now, $\frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 2 x$, And, using the first result from the FTOC:

$\frac{d}{d \left({x}^{2}\right)} {\int}_{1}^{{x}^{2}} \setminus \frac{1}{t} \setminus \mathrm{dt}$ = 1/x^2

Hence combining these trivial results we get:

$\setminus \setminus \setminus \setminus \setminus F ' \left(x\right) = 2 x \cdot \frac{1}{x} ^ 2$
$\therefore F ' \left(x\right) = \frac{2}{x}$