The problem asks to calculate the derivatives of the following three expressions: a) $x^4 - 3x^2 + 4x - 1$ b) $\frac{2x+3}{5x-1}$ c) $\cos(x) \cdot e^x$

1. Problem Description

The problem asks to calculate the derivatives of the following three expressions:

a) $x^4 - 3x^2 + 4x - 1$

b) $\frac{2x+3}{5x-1}$

c) $\cos(x) \cdot e^x$

2. Solution Steps

a) Let $f(x) = x^4 - 3x^2 + 4x - 1$. We need to find $f'(x)$.

Using the power rule and sum/difference rule for derivatives:

$\frac{d}{dx} x^n = nx^{n-1}$

$\frac{d}{dx} [f(x) \pm g(x)] = \frac{d}{dx} f(x) \pm \frac{d}{dx} g(x)$

So, $f'(x) = \frac{d}{dx} (x^4) - 3 \frac{d}{dx} (x^2) + 4 \frac{d}{dx} (x) - \frac{d}{dx} (1)$

$f'(x) = 4x^3 - 3(2x) + 4(1) - 0$

$f'(x) = 4x^3 - 6x + 4$

b) Let $g(x) = \frac{2x+3}{5x-1}$. We need to find $g'(x)$.

Using the quotient rule:

$\frac{d}{dx} \frac{u(x)}{v(x)} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Here, $u(x) = 2x+3$ and $v(x) = 5x-1$.

$u'(x) = \frac{d}{dx} (2x+3) = 2$

$v'(x) = \frac{d}{dx} (5x-1) = 5$

Therefore,

$g'(x) = \frac{2(5x-1) - (2x+3)(5)}{(5x-1)^2}$

$g'(x) = \frac{10x-2 - (10x+15)}{(5x-1)^2}$

$g'(x) = \frac{10x-2 - 10x - 15}{(5x-1)^2}$

$g'(x) = \frac{-17}{(5x-1)^2}$

c) Let $h(x) = \cos(x) \cdot e^x$. We need to find $h'(x)$.

Using the product rule:

$\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

Here, $u(x) = \cos(x)$ and $v(x) = e^x$.

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

$v'(x) = \frac{d}{dx} e^x = e^x$

Therefore,

$h'(x) = -\sin(x)e^x + \cos(x)e^x$

$h'(x) = e^x(\cos(x) - \sin(x))$

3. Final Answer

a) $4x^3 - 6x + 4$

b) $\frac{-17}{(5x-1)^2}$

c) $e^x(\cos(x) - \sin(x))$