If f(x) and g(x) are two functions such that | vedclass

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(D) Given $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$.We know that $(x - \frac{1}{x})^3 = x^3 - \frac{1}{x^3} - 3(x - \frac{1}

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$6$

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(D) Given $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$.We know that $(x - \frac{1}{x})^3 = x^3 - \frac{1}{x^3} - 3(x - \frac{1}{x})$.Therefore,$x^3 - \frac{1}{x^3} = (x - \frac{1}{x})^3 + 3(x - \frac{1}{x})$.Substituting $g(x) = t$,we get $f(t) = t^3 + 3t$.Now,differentiate $f(t)$ with respect to $t$:$f'(t) = \frac{d}{dt}(t^3 + 3t) = 3t^2 + 3$.To find $f'(1)$,substitute $t = 1$:$f'(1) = 3(1)^2 + 3 = 3 + 3 = 6$.

If $f(x)$ and $g(x)$ are two functions such that $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$,then $f'(1)$ is equal to:

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