# Find the derivative of y with respect to theta. y = ln(theta + 1).

## Question:

Find the derivative of {eq}y {/eq} with respect to {eq}\theta {/eq}.

{eq}y = \ln(\theta + 1) {/eq}

## Chain Rule:

Chain rule is applied to determine derivatives of composite functions.

Suppose we are asked to take the derivative of the composite function {eq}y =f(g(x)) {/eq}.

Applying the chain rule:

{eq}y' = f'(g(x))g'(x) {/eq}

Utilize the chain rule to find the derivative of {eq}y {/eq} with respect to {eq}\theta {/eq}:

{eq}\begin{align*} \displaystyle y & = \ln(\theta + 1) \\ \frac{\mathrm{d}y}{\mathrm{d} \theta} & =\frac{D_{\theta}(\theta + 1) }{\theta + 1} \;\;\; \left[ \mathrm{ Employ \ the \ chain \ rule \ and \ the \ common \ derivative \ } D_u \ln u = \displaystyle \frac{\mathrm{d}u}{u}\right]\\ \implies \frac{\mathrm{d}y}{\mathrm{d} \theta} & =\frac{1 }{\theta + 1} \\ \end{align*} {/eq}