Question

F(x)=(x^-2x+1) cosx f'(0)-?

Answer 1

F' (x)= (-2x+1) $x^{-2x}$*(-2)*sinx=(4x+1)* $x^{-2x}$*sinx
F'(0)=1*0*0=0

Answer 2

To find the value of $f'(0)$, we first need to find the derivative of the function $f(x)$ and then evaluate it at $x = 0$.

Given the function $f(x) = (x^{-2x+1})\cos(x)$, we can find its derivative using the product rule and the chain rule.

The product rule states that if $u(x) = g(x) \cdot h(x)$, then $u'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$.

The chain rule states that if $v(x) = u(g(x))$, then $v'(x) = u'(g(x)) \cdot g'(x)$.

Let's apply these rules to find $f'(x)$:

$f(x) = (x^{-2x+1})\cos(x)$ $g(x) = x^{-2x+1}$ $h(x) = \cos(x)$

Now, let's find the derivatives of $g(x)$ and $h(x)$:

$g'(x) = \frac{d}{dx} (x^{-2x+1})$

To differentiate $x^{-2x+1}$, we can use the chain rule and the power rule:

$g'(x) = (1 - 2x) \cdot x^{-2x} \cdot \frac{d}{dx}(x) = (1 - 2x) \cdot x^{-2x} \cdot 1 = (1 - 2x) \cdot x^{-2x}$

$h'(x) = \frac{d}{dx}(\cos(x)) = -\sin(x)$

Now, applying the product rule:

$f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$ $f'(x) = (1 - 2x) \cdot x^{-2x} \cdot \cos(x) + x^{-2x+1} \cdot (-\sin(x))$

Now, we can find $f'(0)$ by evaluating the derivative at $x = 0$:

$f'(0) = (1 - 2 \cdot 0) \cdot 0^{-2 \cdot 0} \cdot \cos(0) + 0^{-2 \cdot 0 + 1} \cdot (-\sin(0))$ $f'(0) = 1 \cdot 1 \cdot 1 + 0 \cdot 0$ $f'(0) = 1$

So, the value of $f'(0)$ is 1.