If f(x) = 5x/(1−x)²/³ + cos² (2x + 1), then f'(0) =

The correct option is 5 – 2 sin 2.

To find $f^{\prime }\left(0\right)$, we need to differentiate the given function $f\left(x\right)$ with respect to $x$ and then evaluate the derivative at $x=0$.

The function is given by:
$$f\left(x\right)=\frac{5x}{{(1-x)}^{2/3}}+{\cos }^2(2x+1)$$

Let us split $f\left(x\right)$ into two parts to differentiate them individually:
Let $u\left(x\right)=\frac{5x}{{(1-x)}^{2/3}}$ and $v\left(x\right)={\cos }^2(2x+1)$.

Step 1: Differentiate $u\left(x\right)$
Using the quotient rule, ${\left(\frac{g}{h}\right)}^{\prime }=\frac{g^{\prime }h-g{h}^{\prime }}{h^2}$:
Here, $g\left(x\right)=5x\Rightarrow g^{\prime }\left(x\right)=5$.
And $h\left(x\right)={(1-x)}^{2/3}\Rightarrow h^{\prime }\left(x\right)=\frac{2}{3}{(1-x)}^{-1/3}·(-1)=-\frac{2}{3}{(1-x)}^{-1/3}$.
Therefore,
$$u^{\prime }\left(x\right)=\frac{5{(1-x)}^{2/3}-5x[-\frac{2}{3}{(1-x)}^{-1/3}]}{{\left[{(1-x)}^{2/3}\right]}^2}$$

Evaluating $u^{\prime }\left(x\right)$ at $x=0$:
$$u^{\prime }\left(0\right)=\frac{5{(1-0)}^{2/3}-5\left(0\right)[-\frac{2}{3}{(1-0)}^{-1/3}]}{{\left[{(1-0)}^{2/3}\right]}^2}=\frac{5\left(1\right)-0}{1}=5$$

Step 2: Differentiate $v\left(x\right)$
Using the chain rule:
$$v^{\prime }\left(x\right)=2\cos (2x+1)·[-\sin (2x+1\left)\right]·2$$
Simplifying using the double-angle identity $2\sin \theta \cos \theta =\sin \left(2\theta \right)$:
$$v^{\prime }\left(x\right)=-2·\left[2\sin \right(2x+1\left)\cos \right(2x+1\left)\right]=-2\sin (4x+2)$$

Evaluating $v^{\prime }\left(x\right)$ at $x=0$:
$$v^{\prime }\left(0\right)=-2\sin \left(4\right(0)+2)=-2\sin 2$$

Step 3: Combine the results
Since $f\left(x\right)=u\left(x\right)+v\left(x\right)$, by the sum rule of differentiation:
$$f^{\prime }\left(0\right)=u^{\prime }\left(0\right)+v^{\prime }\left(0\right)$$
Substituting the calculated values:
$$f^{\prime }\left(0\right)=5-2\sin 2$$