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If $\int_0^x f(t)dt = x^2 + \int_x^1t^2f(t)dt$ , then $f\left(\frac{1}{2} \right )$ is:
Option 1)
$\frac{6}{25}$
Option 2)
$\frac{4}{5}$
Option 3)
$\frac{24}{25}$
Option 4)
$\frac{18}{25}$

Answers (1)

best_answer

Derivative at a point -

The value of f'(x) obtained by putting x = a is called the derivative of f(x) at x = a and it is denoted by f'(a) or

$\frac{dy}{dx}$ at x = a.

-

Algebraic functions -

$\frac{d}{dx}(x^{n})=nx^{n-1}$

$\frac{d}{dx}(\frac{1}{f(x)})^{n}=\frac{-n}{(f(x))^{n+1}}.\frac{d}{dx}f(x)......and\:so\:on$

-

Chain Rule for differentiation (indirect) -

Let y = f(x) is not in standard form then

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$ex:\:\:y=sin(ax+b)$

$Let\:\;u=(ax+b)$

$then\:\:y=sin \:u$

$so\:\:\frac{dy}{du}=cos \:u\:\:and\:\:\frac{du}{dx}=a$

$\therefore \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=a\:cos \:u$

$=a\:cos(ax+b)$

- wherein

$Where\;\:y=f(u)\:\;and\;\;u=f(x)$

$\int_{0}^{x}f(t)dt=x^{2}+\int_{x}^{1}t^{2}f(t)dt$

Differentiate wrt 'x'

$f(x)=2x+0-x^{2}f(x)$

$f(x)=\frac{2x}{x^{2}+1}$

$f{}'(x)=\frac{(1+x^{2})2-(2x)(2x)}{(x^{2}+1)^{2}}$

$=\frac{2x^{2}-4x^{2}+2}{(x^{2}+1)^{2}}$

$f{}'(\frac{1}{2})=\frac{24}{25}$

Option 1)

$\frac{6}{25}$

Option 2)

$\frac{4}{5}$

Option 3)
$\frac{24}{25}$
Option 4)

$\frac{18}{25}$

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