If f(x) = 4x^3 + 3x^2 + 3x + 4,x neq 0,then f | vedclass

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(D) Given $f(x) = 4x^3 + 3x^2 + 3x + 4$. First,find $f\left(\frac{1}{x}\right)$: $f\left(\frac{1}{x}\right) = 4\left(\frac{1}{x}\right)^3 + 3\left(\fr

Vedclass · Accepted Answer

$12x^2 + 6x + 3$

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(D) Given $f(x) = 4x^3 + 3x^2 + 3x + 4$. First,find $f\left(\frac{1}{x}\right)$: $f\left(\frac{1}{x}\right) = 4\left(\frac{1}{x}\right)^3 + 3\left(\frac{1}{x}\right)^2 + 3\left(\frac{1}{x}\right) + 4 = \frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4$. Now,multiply by $x^3$: $x^3 \cdot f\left(\frac{1}{x}\right) = x^3 \left(\frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4\right) = 4 + 3x + 3x^2 + 4x^3$. Finally,differentiate with respect to $x$: $\frac{d}{dx}(4 + 3x + 3x^2 + 4x^3) = 0 + 3 + 6x + 12x^2 = 12x^2 + 6x + 3$. Thus,the correct option is $D$.

If $f(x) = 4x^3 + 3x^2 + 3x + 4$,$x \neq 0$,then $\frac{d}{dx}\left(x^3 \cdot f\left(\frac{1}{x}\right)\right) =$ . . . . . .

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