Question

find the derivative of y with respect to z. y = ln (sinh 2z)

Explanation:

Step1: Apply Chain Rule

Let \( u = \sinh(2z) \), so \( y=\ln(u) \). The chain rule states \( \frac{dy}{dz}=\frac{dy}{du}\cdot\frac{du}{dz} \). First, find \( \frac{dy}{du} \): for \( y = \ln(u) \), \( \frac{dy}{du}=\frac{1}{u} \).

Step2: Differentiate \( u = \sinh(2z) \)

Let \( v = 2z \), so \( u=\sinh(v) \). The derivative of \( \sinh(v) \) with respect to \( v \) is \( \cosh(v) \), and the derivative of \( v = 2z \) with respect to \( z \) is \( 2 \). By the chain rule, \( \frac{du}{dz}=\cosh(v)\cdot2 = 2\cosh(2z) \) (substituting back \( v = 2z \)).

Step3: Combine Results

Substitute \( u=\sinh(2z) \) and \( \frac{du}{dz}=2\cosh(2z) \) into the chain rule formula: \( \frac{dy}{dz}=\frac{1}{\sinh(2z)}\cdot2\cosh(2z) \). Recall that \( \frac{\cosh(x)}{\sinh(x)}=\coth(x) \), so this simplifies to \( 2\coth(2z) \).

Answer:

\( 2\coth(2z) \)