If f(x) = cos x, cos 2 x, cos 4 x, cos 8 x, cos 16 x, then the value of'(π4) is

The correct option is $\sqrt{2}$.

To find the derivative of the given function, we first simplify the product of cosines using a trigonometric identity. The function is given by:
$$f\left(x\right)=cos\left(x\right)cos\left(2x\right)cos\left(4x\right)cos\left(8x\right)cos\left(16x\right)$$

We can multiply and divide the function by $2sin\left(x\right)$:
$$f\left(x\right)=\frac{2sin\left(x\right)cos\left(x\right)cos\left(2x\right)cos\left(4x\right)cos\left(8x\right)cos\left(16x\right)}{2sin\left(x\right)}$$

Using the double-angle identity $2sin\left(\theta \right)cos\left(\theta \right)=sin\left(2\theta \right)$, the product simplifies stage-by-stage:
$2sin\left(x\right)cos\left(x\right)=sin\left(2x\right)$
$2sin\left(2x\right)cos\left(2x\right)=sin\left(4x\right)$
Repeating this process for all terms, we get:
$$f\left(x\right)=\frac{sin\left(32x\right)}{32sin\left(x\right)}$$

Now, we differentiate $f\left(x\right)$ with respect to $x$ using the quotient rule:
$$f^{\text{'}}\left(x\right)=\frac{1}{32}·\frac{32cos\left(32x\right)sin\left(x\right)-sin\left(32x\right)cos\left(x\right)}{{sin}^2\left(x\right)}$$

Next, we evaluate this derivative at $x=\frac{\pi }{4}$:
$sin\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$
$cos\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$
$32x=32\left(\frac{\pi }{4}\right)=8\pi$
$sin\left(8\pi \right)=0$
$cos\left(8\pi \right)=1$

Substitute these values into the derivative expression:
$$f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{1}{32}·\frac{32\left(1\right)\left(\frac{1}{\sqrt{2}}\right)-0}{{\left(\frac{1}{\sqrt{2}}\right)}^2}$$

Simplify the resulting fractions:
$$f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{1}{32}·\frac{\frac{32}{\sqrt{2}}}{\frac{1}{2}}$$

$$f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{1}{32}·\frac{32}{\sqrt{2}}·2$$

$$f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{2}{\sqrt{2}}=\sqrt{2}$$

Therefore, the value of the derivative at the given point is $\sqrt{2}$.