3. a) Hallar la ecuación de la recta tangente a la gráfica de \[ f(x)=\ln (4 x+1)+\frac{7}{3 x+1} \] en \( x=0 \) b) Si \( y=\operatorname{sh}(x) \), verificar que \( y^{\prime \prime}=y \) c) Si \( f(x)=\operatorname{tg}^{2}(x) \) y \( g(x)=\sec ^{2}(x) \), comprobar que \( f^{\prime}(x)=g^{\prime}(x) \)

Alright, I need to solve the given mathematical problems step by step. Let's tackle each part one by one. **Part a) Hallar la ecuación de la recta tangente a la gráfica de** \[ f(x) = \ln(4x + 1) + \frac{7}{3x + 1} \] en \( x = 0 \) First, I need to find the equation of the tangent line to the graph of the function \( f(x) \) at the point where \( x = 0 \). To do this, I'll need to find the slope of the tangent line at that point, which is given by the derivative of \( f(x) \) evaluated at \( x = 0 \). Let's start by finding \( f(0) \): \[ f(0) = \ln(4(0) + 1) + \frac{7}{3(0) + 1} = \ln(1) + \frac{7}{1} = 0 + 7 = 7 \] So, the point of tangency is \( (0, 7) \). Next, I'll find the derivative \( f'(x) \): \[ f(x) = \ln(4x + 1) + \frac{7}{3x + 1} \] Using the chain rule for the first term and the quotient rule for the second term: \[ f'(x) = \frac{4}{4x + 1} - \frac{21}{(3x + 1)^2} \] Now, evaluate \( f'(0) \): \[ f'(0) = \frac{4}{4(0) + 1} - \frac{21}{(3(0) + 1)^2} = 4 - 21 = -17 \] So, the slope of the tangent line at \( x = 0 \) is \( -17 \). Using the point-slope form of a line: \[ y - y_1 = m(x - x_1) \] Plugging in the point \( (0, 7) \) and slope \( m = -17 \): \[ y - 7 = -17(x - 0) \] \[ y = -17x + 7 \] **Part b) Si \( y = \operatorname{sh}(x) \), verificar que \( y^{\prime \prime} = y \)** First, recall that \( \operatorname{sh}(x) \) is the hyperbolic sine function, defined as: \[ \operatorname{sh}(x) = \frac{e^x - e^{-x}}{2} \] Let's find the first and second derivatives of \( y \): \[ y = \operatorname{sh}(x) \] \[ y' = \frac{d}{dx} \left( \frac{e^x - e^{-x}}{2} \right) = \frac{e^x + e^{-x}}{2} = \operatorname{ch}(x) \] \[ y'' = \frac{d}{dx} \left( \frac{e^x + e^{-x}}{2} \right) = \frac{e^x - e^{-x}}{2} = \operatorname{sh}(x) \] So, \( y'' = y \), which verifies the statement. **Part c) Si \( f(x) = \operatorname{tg}^{2}(x) \) y \( g(x) = \sec ^{2}(x) \), comprobar que \( f^{\prime}(x) = g^{\prime}(x) \)** First, let's find the derivatives of both functions. For \( f(x) = \operatorname{tg}^{2}(x) \): \[ f(x) = (\operatorname{tg}(x))^2 \] Using the chain rule: \[ f'(x) = 2 \operatorname{tg}(x) \cdot \operatorname{tg}'(x) \] We know that \( \operatorname{tg}'(x) = \sec^2(x) \), so: \[ f'(x) = 2 \operatorname{tg}(x) \cdot \sec^2(x) \] For \( g(x) = \sec^{2}(x) \): \[ g(x) = (\sec(x))^2 \] Using the chain rule: \[ g'(x) = 2 \sec(x) \cdot \sec(x) \tan(x) = 2 \sec^2(x) \tan(x) \] Now, let's compare \( f'(x) \) and \( g'(x) \): \[ f'(x) = 2 \operatorname{tg}(x) \cdot \sec^2(x) \] \[ g'(x) = 2 \sec^2(x) \tan(x) \] Recall that \( \operatorname{tg}(x) = \frac{\sin(x)}{\cos(x)} \) and \( \sec(x) = \frac{1}{\cos(x)} \), so: \[ \operatorname{tg}(x) = \frac{\sin(x)}{\cos(x)} \] \[ \sec^2(x) = \frac{1}{\cos^2(x)} \] Substituting these into \( f'(x) \): \[ f'(x) = 2 \cdot \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos^2(x)} = 2 \cdot \frac{\sin(x)}{\cos^3(x)} \] Similarly, for \( g'(x) \): \[ g'(x) = 2 \cdot \frac{1}{\cos^2(x)} \cdot \frac{\sin(x)}{\cos(x)} = 2 \cdot \frac{\sin(x)}{\cos^3(x)} \] Therefore, \( f'(x) = g'(x) \), which verifies the statement.